∫ WBCHSE - Important Differential Equation
40+ Differential Equation Marks · 2M 3M
Click on “📋 Click for Solution” button to reveal the answer.
📘 2 Mark Integrals
20+ problemsSolution:
Given,
\(\displaystyle x = A\cos \sqrt{\mu}\, t\)
Differentiating w.r.t. \(t\), we get
\(\displaystyle \frac{dx}{dt} = -(A\sin \sqrt{\mu}\, t)\times \sqrt{\mu}\)
\(\displaystyle \Rightarrow \frac{dx}{dt} = -A\sqrt{\mu}\sin \sqrt{\mu}\, t\)
\(\displaystyle \Rightarrow \frac{dx}{dt} = -\frac{x}{\cos \sqrt{\mu}\, t}\times \sqrt{\mu}\sin \sqrt{\mu}\, t \quad [\text{As } x = A\cos \sqrt{\mu}\, t]\)
\(\displaystyle \therefore \boxed{\frac{dx}{dt} = -x\sqrt{\mu}\tan \sqrt{\mu}\, t}\)
This is the required differential equation.
Solution:
Given,
\(y = ae^{-x} + be^{x} + c\)
Differentiating w.r.t. \(x\), we get
\(\frac{dy}{dx} = -ae^{-x} + be^{x}\)
\(\Rightarrow \frac{dy}{dx} = e^{-x}(-a + be^{2x})\)
\(\Rightarrow e^{x}\frac{dy}{dx} = -a + be^{2x}\)
Differentiating again w.r.t. \(x\), we get
\(e^{x}\frac{d^{2}y}{dx^{2}} + e^{x}\frac{dy}{dx} = 2be^{2x}\)
\(\Rightarrow e^{-x}\frac{d^{2}y}{dx^{2}} + e^{-x}\frac{dy}{dx} = 2b\)
Differentiating again w.r.t. \(x\), we get
\(e^{-x}\frac{d^{3}y}{dx^{3}} - e^{-x}\frac{d^{2}y}{dx^{2}} + e^{-x}\frac{d^{2}y}{dx^{2}} - e^{-x}\frac{dy}{dx} = 0\)
\(\Rightarrow \boxed{\frac{d^{3}y}{dx^{3}} - \frac{dy}{dx} = 0}\)
This is the required differential equation.
Solution:
Given,
\(ax + by + c = 0\)
\(\Rightarrow by = -ax - c\)
\(\Rightarrow y = -\frac{a}{b}x - \frac{c}{b}\)
Differentiating w.r.t. \(x\), we get
\(\frac{dy}{dx} = -\frac{a}{b}\)
Differentiating again, we get
\(\boxed{\frac{d^{2}y}{dx^{2}} = 0}\)
This is the required differential equation.
Solution:
Given,
\(x = ae^{b+t}\)
\(\Rightarrow x = e^{t}(ae^{b}) \quad ...(i)\)
Differentiating w.r.t. \(t\), we get
\(\frac{dx}{dt} = ae^{b}\, e^{t} \quad (ae^{b} \rightarrow \text{constant})\)
\(\Rightarrow \frac{dx}{dt} = \frac{x}{e^{t}} \times e^{t} \quad [\text{Using equation (i)}]\)
\(\therefore \boxed{\frac{dx}{dt} = x}\)
This is the required differential equation.
Solution:
Given,
\(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)
\(\Rightarrow a^{2}y^{2} + b^{2}x^{2} = a^{2}b^{2}\)
\(\Rightarrow y^{2} = -\frac{b^{2}}{a^{2}}x^{2} + b^{2}\)
Differentiating w.r.t. \(x\), we get
\(2y\frac{dy}{dx} = -2x\frac{b^{2}}{a^{2}}\)
\(\Rightarrow y\frac{dy}{dx} = -x\frac{b^{2}}{a^{2}}\)
\(\Rightarrow \frac{y}{x}\frac{dy}{dx} = -\frac{b^{2}}{a^{2}}\)
Differentiating w.r.t. \(x\), we get
\(\left(x\frac{dy}{dx} - y\right)\frac{dy}{dx} - xy\frac{d^{2}y}{dx^{2}} = 0\)
\(\Rightarrow \boxed{x\left(\frac{dy}{dx}\right)^{2} - y\frac{dy}{dx} - xy\frac{d^{2}y}{dx^{2}} = 0}\)
This is the required differential equation.
Solution:
The equations of all circles touching \(x\)-axis at origin are —
\((x-0)^{2} + (y-a)^{2} = a^{2}\), where \(a\) is the length of radius of circle.
So, we have —
\(x^{2} + y^{2} - 2ay + a^{2} = a^{2}\)
\(\Rightarrow x^{2} + y^{2} - 2ay = 0\)
\(\Rightarrow \frac{x^{2} + y^{2}}{y} = 2a\)
Now, differentiating w.r.t. \(x\), we get
\(\frac{y(2x + 2y\frac{dy}{dx}) - (x^{2} + y^{2})\frac{dy}{dx}}{y^{2}} = 0\)
\(\Rightarrow 2xy + 2y^{2}\frac{dy}{dx} - x^{2}\frac{dy}{dx} - y^{2}\frac{dy}{dx} = 0\)
\(\therefore \boxed{y^{2}\frac{dy}{dx} - x^{2}\frac{dy}{dx} + 2xy = 0}\)
This is the required differential equation.
Solution:
Equations of all parabolas whose axis is parallel to the \(x\)-axis are given by —
\((y-\beta)^{2} = 4a(x-\alpha)\), where \((\alpha, \beta)\) are the coordinates of vertex of parabola.
In our case, there are three constants \(a, \alpha\) and \(\beta\).
So, we have to differentiate 3 times.
\((y-\beta)^{2} = 4a(x-\alpha)\)
Differentiating w.r.t. \(x\), we get
\(2(y-\beta)y_{1} = 4a\)
\(\Rightarrow (y-\beta)y_{1} = 2a\)
\(\Rightarrow yy_{1} - \beta y_{1} = 2a\)
Differentiating again w.r.t. \(x\), we get
\((y_{1}^{2} + yy_{2}) - \beta y_{2} = 0 \quad [y_{1}=\frac{dy}{dx},\; y_{2}=\frac{d^{2}y}{dx^{2}}]\)
\(\Rightarrow y + \frac{y_{1}^{2}}{y_{2}} - \beta = 0\)
Differentiating again w.r.t. \(x\), we get
\(y_{1} + \left(\frac{y_{2} \times 2y_{1}y_{2} - y_{1}^{2}y_{3}}{y_{2}^{2}}\right) = 0 \quad [y_{3}=\frac{d^{3}y}{dx^{3}}]\)
\(\Rightarrow y_{1}y_{2}^{2} + 2y_{1}y_{2}^{2} - y_{1}^{2}y_{3} = 0\)
\(\therefore \boxed{y_{1}^{2}y_{3} - 3y_{1}y_{2}^{2} = 0}\)
This is the required differential equation.
Solution:
Given,
\(y = be^{ax}\)
Differentiating w.r.t. \(x\), we get
\(\frac{dy}{dx} = abe^{ax}\)
\(\Rightarrow \frac{dy}{dx} = ay \quad [\because y = be^{ax}]\)
Differentiating again w.r.t. \(x\), we get
\(\frac{d^{2}y}{dx^{2}} = a\frac{dy}{dx}\)
\(\Rightarrow \frac{d^{2}y}{dx^{2}} = \left(\frac{1}{y}\frac{dy}{dx}\right)\frac{dy}{dx} \quad \left[\because a=\frac{1}{y}\frac{dy}{dx}\right]\)
\(\therefore \boxed{y\frac{d^{2}y}{dx^{2}} - \left(\frac{dy}{dx}\right)^{2} = 0}\)
This is the required differential equation of 2nd order and 1st degree.
Solution:
Given,
\(\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{3/2}=p\frac{d^{2}y}{dx^{2}}\)
On squaring both sides, we get
\(\left\{1+\left(\frac{dy}{dx}\right)^{2}\right\}^{3}=p^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{2}\)
\(\therefore \text{Degree} = 2,\quad \text{Order} = 2\)
Solution:
Given,
\(x^{4}\frac{d^{3}y}{dx^{3}} + x^{2}\frac{dy}{dx} = 2\)
\(\therefore \text{Degree} = 1,\quad \text{Order} = 3\)
Solution:
Given,
\(\left(\frac{d^{2}y}{dt^{2}} + t\frac{dy}{dt}\right)^{1/2} = \sqrt[3]{1+\frac{dy}{dt}} = \left(1+\frac{dy}{dt}\right)^{1/3}\)
Now, both sides raised to power 6, we get
\(\left(\frac{d^{2}y}{dt^{2}} + t\frac{dy}{dt}\right)^{3} = \left(1+\frac{dy}{dt}\right)^{2}\)
\(\Rightarrow \left(\frac{d^{2}y}{dt^{2}}\right)^{3} + 3\left(\frac{d^{2}y}{dt^{2}}\right)^{2} t\frac{dy}{dt} + 3t^{2}\frac{d^{2}y}{dt^{2}}\left(\frac{dy}{dt}\right)^{2} + t^{3}\left(\frac{dy}{dt}\right)^{3} = \left(1+\frac{dy}{dt}\right)^{2}\)
\(\therefore \text{Degree} = 3,\quad \text{Order} = 2\)
Solution:
Given,
\((1+y^{2})dx = xy\,dy\)
\(\Rightarrow \frac{dx}{x} = \frac{y}{1+y^{2}}\,dy\)
Integrating both sides, we get
\(\int \frac{dx}{x} = \frac{1}{2}\int \frac{2y}{1+y^{2}}\,dy + \ln C\) (\(\ln C\) = Constant of integration)
\(\Rightarrow \ln|x| = \frac{1}{2}\ln|1+y^{2}| + \ln C\)
\(\Rightarrow 2\ln\left|\frac{x}{C}\right| = \ln|1+y^{2}|\)
\(\Rightarrow \left(\frac{x}{C}\right)^{2} = (1+y^{2})\)
\(\Rightarrow x^{2} = C^{2}(1+y^{2})\)
Now, the above curve passes through \((1,0)\), so \(x=1,\; y=0\)
Therefore, \( 1^{2} = C^{2}(1+0^{2})\)
\(\Rightarrow C^{2} = 1\)
Hence, the required equation of the curve is \(x^{2} = 1+y^{2}\).
Solution:
Given,
\(\frac{dy}{dx} = \frac{1+y^{2}}{xy}\)
\(\Rightarrow \frac{1}{2}\left(\frac{2y}{1+y^{2}}\right)dy = \frac{dx}{x}\)
Integrating both sides, we get
\(\frac{1}{2}\int \frac{2y}{1+y^{2}}\,dy = \int \frac{dx}{x} + \ln|C| \quad (\ln|C| = \text{constant of integration})\)
\(\Rightarrow \frac{1}{2}\ln|1+y^{2}| = \ln|x| + \ln|C|\)
\(\Rightarrow \frac{1}{2}\ln|1+y^{2}| = \ln|Cx|\)
\(\Rightarrow \ln|1+y^{2}| = 2\ln|Cx|\)
\(\Rightarrow 1+y^{2} = (Cx)^{2} = C^{2}x^{2} \quad ...(i)\)
As given \(y(1)=0\) i.e. at \(x=1,\; y=0\)
\(\Rightarrow 1+0^{2} = C^{2}\cdot 1^{2}\)
\(\therefore C^{2}=1\)
Therefore, the required solution to the given differential equation is
\(\boxed{1+y^{2}=x^{2}}\)
Solution:
Given,
\(e^{y-x}dx - e^{x-y}dy = 0\)
\(\Rightarrow e^{x-y}dy = e^{y-x}dx\)
\(\Rightarrow \frac{e^{x-y}}{e^{y-x}}dy = dx\)
\(\Rightarrow e^{2x-2y}dy = dx\)
\(\Rightarrow e^{-2y}dy = e^{-2x}dx\)
Integrating both sides, we get
\(\int e^{-2y}dy = \int e^{-2x}dx + C^{'}\)
\(\Rightarrow \frac{e^{-2y}}{-2} = \frac{e^{-2x}}{-2} + C^{'}\)
\(\Rightarrow e^{-2y} = e^{-2x} + C\)
\((C = 2C^{'} \text{constant of integration})\)
Therefore, the required solution to the given differential equation is \(e^{-2y} = e^{-2x} + C \)
Solution:
Given,
\(x\frac{dy}{dx} - y = x^{2}\)
\(\Rightarrow \frac{dy}{dx} + \left(-\frac{1}{x}\right)y = x \quad (x \neq 0)\)
On comparing the above equation with \(\frac{dy}{dx} + Py = Q\), we get
\(P = -\frac{1}{x}\)
Therefore, I.F. \(= e^{\int P\,dx}\)
\(= e^{\int -\frac{1}{x}\,dx}\)
\(= e^{-\ln|x|}\)
\(= e^{\ln\left|\frac{1}{x}\right|} \, \, [\because -ln \, |\frac{1}{x}| = ln \, x]\)
\(= \frac{1}{x} \, \, [\because e^{ln \, x} = x]\)
Therefore, the required (I.F.) is \(\frac{1}{x}\)
Solution:
Given,
\(\frac{dy}{dx} + y\tan x = \sec x\)
Compare it with the following differential equation —
\(\frac{dy}{dx} + Py = Q\), we get
\(P = \tan x\)
Now, I.F. \(= e^{\int P\,dx}\)
\(= e^{\int \tan x\,dx}\)
\(= e^{\ln|\sec x|}\quad (\because \int \tan x\,dx = \ln|\sec x| + C)\)
\(= \sec x \, \, [\because e^{ln \, x} = x]\)
Hence, I.F. \(= \sec x\).
Solution:
As, the slope of tangent i.e. \(\frac{dy}{dx}\) at any point on the curve is equal to its reciprocal of ordinate.
\(\therefore \frac{dy}{dx} = \frac{1}{y}\)
\(\Rightarrow y\,dy = dx\)
Now integrating both sides, we get
\(\int y\,dy = \int dx + C\)
\(\Rightarrow \frac{y^{2}}{2} = x + C\)
It passes through \((4,3)\), so
\(\Rightarrow \frac{3^{2}}{2} = 4 + C\)
\(\Rightarrow C = \frac{9}{2} - 4 = \frac{1}{2}\)
Hence, the required equation of curve is
\(\boxed{y^{2} = 2x + 1}\)
Solution:
Given,
\((x+y)(dx-dy)=dx+dy\)
\(\Rightarrow xdx - xdy + ydx - ydy - dx - dy = 0\)
\(\Rightarrow (x-1)dx - (y+1)dy = xdy + ydx\)
\(\Rightarrow (x-1)dx - (y+1)dy = d(xy)\)
Now integrating both sides, we get
\(\int (x-1)dx - \int (y+1)dy = \int d(xy) + C'\)
\(\Rightarrow \frac{x^{2}}{2} - x - \frac{y^{2}}{2} - y = xy + C'\)
\(\Rightarrow x^{2} - y^{2} - 2x - 2y = 2xy + C\)
\((C=2C' \text{, constant of integration})\)
Solution:
Given,
\(x\sqrt{1+y^{2}}\,dx + y\sqrt{1+x^{2}}\,dy = 0\)
\(\Rightarrow \frac{x\,dx}{\sqrt{1+x^{2}}} + \frac{y\,dy}{\sqrt{1+y^{2}}} = 0\)
Integrating both sides, we get
\(\int \frac{x\,dx}{\sqrt{1+x^{2}}} + \int \frac{y\,dy}{\sqrt{1+y^{2}}} = C\)
Let, \(1+x^{2} = z^{2}\)
Differentiating w.r.t. \(x\), we get
\(2x\,dx = 2z\,dz\)
\(\Rightarrow x\,dx = z\,dz\)
Similarly, let \(1+y^{2} = t^{2}\)
\(\Rightarrow y\,dy = t\,dt\)
So, the differential equation becomes,
\(\int \frac{z\,dz}{z} + \int \frac{t\,dt}{t} = C\)
\(\Rightarrow z + t = C\)
\(\therefore \boxed{\sqrt{1+x^{2}} + \sqrt{1+y^{2}} = C} \, \, \,\) - This is the required solution of the given differential equation.
Solution:
Given,
\(\log\left(\frac{dy}{dx}\right) = 4x - 5y\)
\(\Rightarrow \frac{dy}{dx} = e^{4x-5y}\quad (\because \log a = c \Rightarrow e^{c}=a)\)
\(\Rightarrow e^{5y}dy = e^{4x}dx\)
Integrating both sides, we get
\(\int e^{5y}dy = \int e^{4x}dx + C'\)
\(\Rightarrow \frac{e^{5y}}{5} = \frac{e^{4x}}{4} + C'\)
\(\therefore \boxed{4e^{5y} = 5e^{4x} + C} \quad ....\) This is the required solution of the given differential equation.
Where \((C = 20C', \text{constant of integration})\)
Solution:
Given,
\(y = a\cos(\log x) + b\sin(\log x) \quad ...(i)\)
Differentiating w.r.t. \(x\), we get
\(\frac{dy}{dx} = \frac{-a\sin(\log x) + b\cos(\log x)}{x}\)
\(\Rightarrow x\frac{dy}{dx} = -a\sin(\log x) + b\cos(\log x)\)
Differentiating w.r.t. \(x\) again, we get
\(x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = \frac{-a\cos(\log x) - b\sin(\log x)}{x}\)
\(\Rightarrow x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} = -(a\cos(\log x) + b\sin(\log x))\)
\(\Rightarrow x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + y = 0 \quad [\text{Using equation (i)}]\)
Hence, \(x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + y = 0\) is satisfied.
Solution:
Given,
\(x = e^{-t}(a\cos t + b\sin t) \quad ...(i)\)
Differentiating w.r.t. \(t\), we get
\(\frac{dx}{dt} = -e^{-t}(a\cos t + b\sin t) + e^{-t}(-a\sin t + b\cos t)\)
\(\Rightarrow \frac{dx}{dt} = -x + e^{-t}(-a\sin t + b\cos t) \quad \text{(Using equation (i))} \quad ...(ii)\)
Differentiating again w.r.t. \(t\), we get
\(\frac{d^{2}x}{dt^{2}} = -\frac{dx}{dt} + \left[-e^{-t}(-a\sin t + b\cos t) + e^{-t}(-a\cos t - b\sin t)\right]\)
\(\Rightarrow \frac{d^{2}x}{dt^{2}} = -\frac{dx}{dt} - e^{-t}(-a\sin t + b\cos t) - e^{-t}(a\cos t + b\sin t)\)
\(\Rightarrow \frac{d^{2}x}{dt^{2}} = -\frac{dx}{dt} - \left(\frac{dx}{dt} + x\right) - x \quad \text{[Using equation (i) and (ii)]}\)
\(\Rightarrow \frac{d^{2}x}{dt^{2}} = -2\frac{dx}{dt} - 2x\)
\(\therefore \boxed{\frac{d^{2}x}{dt^{2}} + 2\frac{dx}{dt} + 2x = 0} \quad ... \) This is the required differential equation.
Solution:
Given,
\(x^{2}(4+y^{2})dx + y^{2}(4+x^{2})dy = 0\)
\(\Rightarrow \frac{x^{2}}{4+x^{2}}dx + \frac{y^{2}}{4+y^{2}}dy = 0\)
Integrating both sides, we get
\(\int \frac{x^{2}}{4+x^{2}}dx + \int \frac{y^{2}}{4+y^{2}}dy = C\)
\(\Rightarrow \int \frac{(4+x^{2})-4}{4+x^{2}}dx + \int \frac{(4+y^{2})-4}{4+y^{2}}dy = C\)
\(\Rightarrow \int dx - 4\int \frac{dx}{4+x^{2}} + \int dy - 4\int \frac{dy}{4+y^{2}} = C\)
\(\Rightarrow x - 4\left(\frac{1}{2}\tan^{-1}\frac{x}{2}\right) + y - 4\left(\frac{1}{2}\tan^{-1}\frac{y}{2}\right) = C\)
\(\therefore \boxed {x + y + 2\left(\tan^{-1}\frac{x}{2} + \tan^{-1}\frac{y}{2}\right) = C} \quad ... \) This is the required solution of the given differential equation.
📗 3 Mark Integrals
15+ problemsSolution:
Given,
\(\log\left(\frac{dy}{dx}\right) = 3x - 5y\)
\(\Rightarrow \frac{dy}{dx} = e^{3x-5y}\)
\(\Rightarrow e^{5y}dy = e^{3x}dx\)
Integrating both sides, we get
\(\int e^{5y}dy = \int e^{3x}dx + C\)
\(\Rightarrow \frac{e^{5y}}{5} = \frac{e^{3x}}{3} + C\)
\(\Rightarrow 3e^{5y} = 5e^{3x} + C\)
This is the required solution.
Solution:
Given,
\(x\frac{dy}{dx} - y = x\tan\left(\frac{y}{x}\right)\)
\(\Rightarrow \frac{dy}{dx} - \frac{y}{x} = \tan\left(\frac{y}{x}\right) \quad ..... (i)\)
Let, \(v=\frac{y}{x}\) ⇒ \(y=vx\)
\(\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\)
Substituting \(\frac{dy}{dx} = v + x\frac{dv}{dx} \, \text{and} \, \frac{y}{x}=v \) in the equation \((i) \), we get
\(v + x\frac{dv}{dx} - v = \tan v\)
\(\Rightarrow x\frac{dv}{dx} = \tan v\)
\(\Rightarrow \frac{dv}{\tan v} = \frac{dx}{x}\)
\(\Rightarrow \int \cot v\, dv = \int \frac{dx}{x}\)
\(\Rightarrow \ln|\sin v| = \ln|x| + C^{'}\)
\(\Rightarrow \ln|\sin v| = \ln|x| + ln \,C\)
\(\Rightarrow \sin v = Cx\)
\(\Rightarrow \sin\left(\frac{y}{x}\right) = Cx \quad .... (ii)\)
Given, \(y(1)=\frac{\pi}{2}\), so using \( x=1 \, \text{and} \, y=\frac{\pi}{2}\), we get
\(\Rightarrow \sin\left(\frac{\pi}{2}\right) = C(1)\)
\(\Rightarrow 1 = C\)
Therefore, from equation \((i)\), we will get \(\boxed{\sin\left(\frac{y}{x}\right) = x}\)
Solution:
Given,
\(\frac{dy}{dx} = (1+y^{2})\left(x+\tan^{-1}y\right)\)
\(\Rightarrow dy = (1+y^{2})\left(x+\tan^{-1}y\right)dx \quad ...(i)\)
Let, \(v=\tan^{-1}y\) ⇒ \(y=\tan v\)
\(\Rightarrow dv = \frac{1}{1+y^{2}} \, dy \)
\(\Rightarrow dy = (1+y^{2}) \, dv \)
Substituting \(dy = (1+y^{2}) \, dv \) in equation \((i)\), we get
\((1+y^{2})\, dv = (1+y^{2})(x+v)\, dx\)
\(\Rightarrow \frac{dv}{dx} = x+v\)
\(\Rightarrow \frac{dv}{dx} - v = x \quad ...(ii)\)
This is a linear differential equation.
I.F. \(= e^{\int -1\,dx} = e^{-x}\)
Multiplying equation \((ii)\) by I.F., we get
\(e^{-x}\frac{dv}{dx} - ve^{-x} = xe^{-x}\)
\(\Rightarrow \frac{d}{dx}(ve^{-x}) = xe^{-x}\)
Integrating both sides, we get
\(ve^{-x} = \int xe^{-x}dx + C\)
\(\Rightarrow ve^{-x}= - (x+1)e^{-x} + C\)
\(\Rightarrow v = -(x+1) + Ce^{x}\) (Here integration by parts is used for \(\int xe^{-x}dx\) which equals \(- (x+1)e^{-x} \))
\(\Rightarrow \tan^{-1}y = Ce^{x} - x - 1\)
\(\therefore \boxed{\tan^{-1}y + x + 1 = Ce^{x}}\)
Solution:
Given,
\(ydx - xdy + 3x^{2}y^{2}e^{x^{3}}dx = 0\)
\(\Rightarrow ydx - xdy = -3x^{2}y^{2}e^{x^{3}}dx \quad ...(i)\)
We know that
\(d\!\left(\frac{x}{y}\right) = \frac{y\,dx - x\,dy}{y^{2}}\)
\(\Rightarrow ydx - xdy = y^{2} d\!\left(\frac{x}{y}\right)\)
Substituting in equation \((i)\), we get
\(y^{2} d\!\left(\frac{x}{y}\right) = -3x^{2}y^{2}e^{x^{3}}dx\)
\(\Rightarrow d\!\left(\frac{x}{y}\right) = -3x^{2}e^{x^{3}}dx\)
Integrating both sides, we get
\(\int d\!\left(\frac{x}{y}\right) = -\int 3x^{2}e^{x^{3}}dx + C\)
\(\Rightarrow \frac{x}{y} = -e^{x^{3}} + C\)
\(\therefore \boxed{\frac{x}{y} + e^{x^{3}} = C}\)
Solution:
Given,
\((x^{2}+1)\frac{dy}{dx} + 2xy = \sqrt{x^{2}+4}\)
\(\Rightarrow \frac{dy}{dx} + \frac{2x}{x^{2}+1}y = \frac{\sqrt{x^{2}+4}}{x^{2}+1} \quad ...(i)\)
This is a linear differential equation.
\(P=\frac{2x}{x^{2}+1}\)
I.F. \(= e^{\int \frac{2x}{x^{2}+1}dx}\)
\(= e^{\ln(x^{2}+1)}\)
\(= x^{2}+1\)
Multiplying equation \((i)\) by I.F., we get
\((x^{2}+1)\frac{dy}{dx} + 2xy = \sqrt{x^{2}+4}\)
\(\Rightarrow \frac{d}{dx}\left[(x^{2}+1)y\right] = \sqrt{x^{2}+4}\)
\(\Rightarrow d \left[(x^{2}+1)y \right] = \sqrt{x^{2}+4} \, dx\)
Integrating both sides, we get
\((x^{2}+1)y = \int \sqrt{x^{2}+4}\,dx + C\)
Using the standard result,
\(\int \sqrt{x^{2}+a^{2}}\,dx = \frac{x}{2}\sqrt{x^{2}+a^{2}} + \frac{a^{2}}{2}\ln\left|x+\sqrt{x^{2}+a^{2}}\right|\)
Here, \(a^{2}=4\), so
\(\int \sqrt{x^{2}+4}\,dx = \frac{x}{2}\sqrt{x^{2}+4} + 2\ln\left|x+\sqrt{x^{2}+4}\right|\)
\(\Rightarrow (x^{2}+1)y = \frac{x}{2}\sqrt{x^{2}+4} + 2\ln\left|x+\sqrt{x^{2}+4}\right| + C\)
\(\therefore \boxed{y = \frac{\frac{x}{2}\sqrt{x^{2}+4} + 2\ln\left|x+\sqrt{x^{2}+4}\right| + C}{x^{2}+1}}\)
Solution:
Given,
\(\frac{dy}{dx} = \frac{\cos(\log x)}{\log y}\)
\(\Rightarrow \log y \, dy = \cos(\log x)\, dx \)
Integrating both sides, we get
\(\int \log y \, dy = \int \cos(\log x)\, dx + C \quad ...(i)\)
\(\int \log y \, dy = y\log y - y\)
For \(\int \cos(\log x)\, dx\), let \(t=\log x\)
\(\Rightarrow dt=\frac{1}{x}dx \quad \Rightarrow dx=e^{t}dt\)
\(\Rightarrow \int \cos(\log x)\, dx = \int e^{t}\cos t \, dt\)
Let \(I=\int e^{t}\cos t \, dt\)
\(I=\cos t \int e^{t}dt - \int \left(\frac{d}{dt}\cos t\right)\left(\int e^{t}dt\right)dt\)
\(=e^{t}\cos t + \int e^{t}\sin t\,dt\)
\(=e^{t}\cos t + \sin t\int e^{t}dt - \int \left(\frac{d}{dt}\sin t\right)\left(\int e^{t}dt\right)dt\)
\(=e^{t}\cos t + e^{t}\sin t - \int e^{t}\cos t\,dt\)
\(=e^{t}\cos t + e^{t}\sin t - I\)
\(\Rightarrow 2I=e^{t}(\sin t+\cos t)\)
\(\Rightarrow I=\frac{e^{t}}{2}(\sin t+\cos t)\)
Substituting \(t=\log x\),
\(\int \cos(\log x)\, dx=\frac{x}{2}\left[\sin(\log x)+\cos(\log x)\right]\)
Therefore, from equation \((i)\),
\(\boxed{y\log y - y=\frac{x}{2}\left[\sin(\log x)+\cos(\log x)\right]+C}\)
Solution:
Given,
\(\frac{dy}{dx} = (x-y)^{6} + 1 \quad ...(i)\)
Let, \(v = x-y\)
\(\Rightarrow \frac{dv}{dx} = 1 - \frac{dy}{dx}\)
Substituting \(\frac{dy}{dx} = (x-y)^{6} + 1 = v^{6}+1\) in above, we get
\(\frac{dv}{dx} = 1 - (v^{6}+1)\)
\(\Rightarrow \frac{dv}{dx} = -v^{6}\)
\(\Rightarrow \frac{dv}{v^{6}} = -dx\)
\(\Rightarrow \int v^{-6} dv = -\int dx + C\)
\(\Rightarrow \frac{v^{-5}}{-5} = -x + C\)
\(\Rightarrow \frac{1}{5v^{5}} = x + C\)
Substituting \(v = x-y\),
\(\therefore \boxed{\frac{1}{5(x-y)^{5}} = x + C}\)
Solution:
Let \(N\) be the number of bacteria at time \(t\).
Given,
\(\frac{dN}{dt} \propto N^{1/3}\)
\(\Rightarrow \frac{dN}{dt} = kN^{1/3} \quad ...(i)\)
\(\Rightarrow \frac{dN}{N^{1/3}} = k\,dt\)
Integrating both sides w.r.t., we get
\(\Rightarrow \int N^{-1/3} dN = \int k\,dt + C\)
\(\Rightarrow \frac{N^{2/3}}{2/3} = kt + C\)
\(\Rightarrow \frac{3}{2}N^{2/3} = kt + C\)
Let initial number be \(N_0\) at \(t=0\).
\(\Rightarrow \frac{3}{2}N_0^{2/3} = C\)
Given, number becomes \(8N_0\) in 3 hours.
\(\Rightarrow \frac{3}{2}(8N_0)^{2/3} = 3k + \frac{3}{2}N_0^{2/3}\)
\(\Rightarrow \frac{3}{2}(4N_0^{2/3}) = 3k + \frac{3}{2}N_0^{2/3}\) [\(\because (8)^{2/3} = 4\)]
\(\Rightarrow 6N_0^{2/3} = 3k + \frac{3}{2}N_0^{2/3}\)
\(\Rightarrow 3k = \frac{9}{2}N_0^{2/3}\)
\(\Rightarrow k = \frac{3}{2}N_0^{2/3}\)
Now for \(N=64N_0\),
\(\Rightarrow \frac{3}{2}(16N_0^{2/3}) = kt + \frac{3}{2}N_0^{2/3}\) [\(\because (64)^{2/3} = 16\)]
\(\Rightarrow 24N_0^{2/3} = \frac{3}{2}N_0^{2/3}t + \frac{3}{2}N_0^{2/3}\)
\(\Rightarrow 24 = \frac{3}{2}(t+1)\)
\(\Rightarrow 48 = 3(t+1)\)
\(\Rightarrow 16 = t+1\)
\(\Rightarrow t=15\)
\(\therefore\) The number will be 64 times in 15 hours.
Solution:
Given,
\(x\frac{dy}{dx} - y = -x\sin\left(\frac{y}{x}\right)\)
\(\Rightarrow \frac{dy}{dx} - \frac{y}{x} = -\sin\left(\frac{y}{x}\right) \quad ...(i)\)
Let, \(v=\frac{y}{x}\) ⇒ \(y=vx\)
\(\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\)
Substituting in equation \((i)\), we get
\(v + x\frac{dv}{dx} - v = -\sin v\)
\(\Rightarrow x\frac{dv}{dx} = -\sin v\)
\(\Rightarrow \frac{dv}{\sin v} = -\frac{dx}{x}\)
\(\Rightarrow \int \csc v\, dv = -\int \frac{dx}{x}\)
\(\Rightarrow \ln\left|\tan\frac{v}{2}\right| = -\ln|x| + C\)
\(\Rightarrow \ln\left|\tan\frac{v}{2}\right| = \ln\left(\frac{C}{x}\right)\)
\(\Rightarrow \tan\frac{v}{2} = \frac{C}{x}\)
\(\Rightarrow \tan\left(\frac{y}{2x}\right) = \frac{C}{x} \quad ...(ii)\)
Given, \(y=\pi\) when \(x=2\)
\(\Rightarrow \tan\left(\frac{\pi}{4}\right) = \frac{C}{2}\)
\(\Rightarrow 1 = \frac{C}{2}\)
\(\Rightarrow C=2\)
Therefore, from equation \((ii)\),
\(\boxed{\tan\left(\frac{y}{2x}\right) = \frac{2}{x}}\)
Solution:
Given,
\(x\log x \frac{dy}{dx} + y = \frac{2}{x}\log x\)
\(\Rightarrow \frac{dy}{dx} + \frac{1}{x\log x}y = \frac{2}{x^{2}}\quad ...(i)\)
This is a linear differential equation.
\(P=\frac{1}{x\log x}\)
I.F. \(= e^{\int \frac{1}{x\log x}dx}\)
\(= e^{\log(\log x)}\)
\(= \log x\)
Multiplying equation \((i)\) by I.F., we get
\(\log x \frac{dy}{dx} + \frac{1}{x}y = \frac{2\log x}{x^{2}}\)
\(\Rightarrow \frac{d}{dx}(y\log x) = \frac{2\log x}{x^{2}}\)
Integrating both sides, we get
\(y\log x = \int \frac{2\log x}{x^{2}}dx + C \quad ....(ii)\)
Let \(t=\log x\)
\(\Rightarrow dt=\frac{1}{x}dx\)
\(\Rightarrow \int \frac{2\log x}{x^{2}}dx = 2\int t e^{-t}dt\)
Using integration by parts,
\(\int t e^{-t}dt \)
\(=t \int e^{-t}dt - \int \left[t \int e^{-t}dt \, \right ] dt \)
\(=-t \, e^{-t} + \int e^{-t}dt \)
\(=-t \, e^{-t} - e^{-t} \)
\(= -(t+1)e^{-t}\)
\(\Rightarrow 2\int t e^{-t}dt = -2(t+1)e^{-t}\)
\(= -\frac{2(\log x +1)}{x}\) putting it into equation \((ii) \), we get
\(\Rightarrow y\log x = -\frac{2(\log x +1)}{x} + C\)
\(\therefore \boxed{y = \frac{C}{\log x} - \frac{2(\log x +1)}{x\log x}}\)
Solution:
Given,
\(\frac{dy}{dx} + y\tan x = 2x + x^{2}\tan x\)
\(\Rightarrow \frac{dy}{dx} + (\tan x)y = (2x + x^{2}\tan x)\)
This is a first order linear differential equation of the form
\(\frac{dy}{dx} + Py = Q\)
Hence \(P=\tan x\) and \(Q=2x + x^{2}\tan x\)
The integrating factor (I.F.) is —
\(\text{I.F.} = e^{\int P\,dx}\)
\(= e^{\int \tan x\,dx}\)
\(= e^{\log(\sec x)}\)
\(= \sec x \quad [\text{As } e^{\log x}=x]\)
Hence, the solution of the differential equation is —
\(y(\text{I.F.}) = \int (Q \times \text{I.F.})\,dx + C\)
\(\Rightarrow y\sec x = \int (2x + x^{2}\tan x)\sec x\,dx + C\)
\(\Rightarrow y\sec x = 2\int x\sec x\,dx + \int x^{2}\sec x\tan x\,dx + C\)
\(\Rightarrow y\sec x = 2\left[ \sec x \int x\,dx - \int \left\{\frac{d}{dx}(\sec x)\right\}\left(\int x\,dx\right) dx \right] + \int x^{2}\sec x\tan x\,dx + C\)
\(\Rightarrow y\sec x = 2\left[ \frac{x^{2} \times \sec x }{2} - \int \sec x\tan x \times \frac{x^{2}}{2} dx \right] + \int x^{2}\sec x\tan x\,dx + C\)
\(\Rightarrow y\sec x = 2\left[ \frac{x^{2}}{2}\sec x - \frac{1}{2}\int x^{2}\sec x\tan x\,dx \right] + \int x^{2}\sec x\tan x\,dx + C\)
\(\Rightarrow y\sec x = x^{2}\sec x - \int x^{2}\sec x\tan x\,dx + \int x^{2}\sec x\tan x\,dx + C\)
\(\Rightarrow y\sec x = x^{2}\sec x + C \quad ...(i)\)
Given, \(y=0\) at \(x=0\), so
\(0 = 0 + C \Rightarrow C=0\)
Hence, the required solution is
\(y\sec x = x^{2}\sec x \Rightarrow \boxed{y=x^{2}}\)
Solution:
Given,
\((1+x^{3})\frac{dy}{dx} + 6x^{2}y = 1+x^{2}\)
\(\Rightarrow \frac{dy}{dx} + \frac{6x^{2}}{1+x^{3}}y = \frac{1+x^{2}}{1+x^{3}}\)
This is a first order linear differential equation of the form
\(\frac{dy}{dx} + Py = Q\)
Here, \(P=\frac{6x^{2}}{1+x^{3}}\) and \(Q=\frac{1+x^{2}}{1+x^{3}}\)
The integrating factor (I.F.) is —
\(\text{I.F.} = e^{\int P\,dx}\)
\(= e^{\int \frac{6x^{2}}{1+x^{3}}dx}\)
Let \(t=1+x^{3}\)
\(\Rightarrow dt=3x^{2}dx\)
\(\Rightarrow \int \frac{6x^{2}}{1+x^{3}}dx = 2\int \frac{dt}{t}\)
\(=2\log t\)
\(=2\log(1+x^{3})\)
\(\Rightarrow \text{I.F.} = e^{2\log(1+x^{3})}\)
\(=(1+x^{3})^{2}\)
Hence, the solution is —
\(y(\text{I.F.}) = \int (Q \times \text{I.F.})dx + C\)
\(\Rightarrow y(1+x^{3})^{2} = \int \frac{1+x^{2}}{1+x^{3}}(1+x^{3})^{2}dx + C\)
\(\Rightarrow y(1+x^{3})^{2} = \int (1+x^{2})(1+x^{3})dx + C\)
\(\Rightarrow y(1+x^{3})^{2} = \int (1 + x^{2} + x^{3} + x^{5})dx + C\)
\(\therefore y(1+x^{3})^{2} = x + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \frac{x^{6}}{6} + C \quad ....(i)\)
Now use the given condition \(y=1\) when \(x=1\).
\(1(1+1)^{2} = 1 + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + C\)
\(4 = 1 + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + C\)
\(4 = \frac{12+4+3+2}{12} + C\)
\(4 = \frac{21}{12} + C\)
\(4 = \frac{7}{4} + C\)
\(\Rightarrow C = \frac{9}{4}\)
Therefore, from equation \((i)\), we get \( \boxed{y = \frac{x + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \frac{x^{6}}{6} + \frac{9}{4}}{(1+x^{3})^{2}}}\)
Solution:
Given,
\(\frac{dy}{dx} = 1 + e^{2x-y} \quad ...(i)\)
\(\Rightarrow \frac{dy}{dx} - 1 = e^{2x-y}\)
\(\Rightarrow \frac{dy}{dx} - 1 = e^{2x}e^{-y}\)
Let \(v = e^{y}\)
\(\Rightarrow \frac{dv}{dx} = e^{y}\frac{dy}{dx} = v\frac{dy}{dx}\)
\(\Rightarrow \frac{dy}{dx} = \frac{1}{v}\frac{dv}{dx}\)
Substituting in equation (i), we get
\(\frac{1}{v}\frac{dv}{dx} = 1 + \frac{e^{2x}}{v}\)
\(\Rightarrow \frac{dv}{dx} = v + e^{2x}\)
\(\Rightarrow \frac{dv}{dx} - v = e^{2x}\)
This is a first order linear differential equation.
\(P=-1\)
I.F. \(= e^{\int -1\,dx} = e^{-x}\)
Multiplying by I.F.,
\(e^{-x}\frac{dv}{dx} - ve^{-x} = e^{2x}e^{-x}\)
\(\Rightarrow \frac{d}{dx}(ve^{-x}) = e^{x}\)
Integrating both sides,
\(ve^{-x} = \int e^{x}dx + C\)
\(\Rightarrow ve^{-x} = e^{x} + C\)
\(\Rightarrow v = e^{2x} + Ce^{x}\)
\(\Rightarrow e^{y} = e^{2x} + Ce^{x} \quad ....(i)\)
Using \(y(2)=2\),
\(e^{2} = e^{4} + Ce^{2}\)
\(\Rightarrow 1 = e^{2} + C\)
\(\Rightarrow C = 1 - e^{2}\)
Therefore, from equation \((i)\), we get \( \boxed{e^{y} = e^{2x} + (1-e^{2})e^{x}}\)
Solution:
Let the population at time \(t\) be \(P\).
Rate of increase is proportional to the population.
\(\Rightarrow \frac{dP}{dt} = 0.05P\)
\(\Rightarrow \frac{dP}{P} = 0.05\,dt\)
Integrating both sides,
\(\int \frac{dP}{P} = \int 0.05\,dt\)
\(\Rightarrow \ln P = 0.05t + C\)
\(\Rightarrow P = Ce^{0.05t}\)
Let initial population be \(P_0\) at \(t=0\).
\(\Rightarrow P_0 = C\)
\(\Rightarrow P = P_0 e^{0.05t}\)
Population doubles when \(P = 2P_0\).
\(\Rightarrow 2P_0 = P_0 e^{0.05t}\)
\(\Rightarrow 2 = e^{0.05t}\)
\(\Rightarrow \ln 2 = 0.05t\)
\(\Rightarrow t = \frac{\ln 2}{0.05}\)
\(\Rightarrow t = 20\ln 2\)
\(\Rightarrow t \approx 13.86 \text{ years}\)
\(\therefore \boxed{\text{Population will double in } 20\ln 2 \text{ years } (\approx 13.86 \text{ years})}\)
Solution:
Given,
\((xy-1)\frac{dy}{dx} + y^{2} = 0\)
\(\Rightarrow (xy-1)\frac{dy}{dx} = -y^{2}\)
\(\Rightarrow \frac{dx}{dy} = -\frac{xy-1}{y^{2}}\)
\(\Rightarrow \frac{dx}{dy} = -\frac{x}{y} + \frac{1}{y^{2}}\)
\(\Rightarrow \frac{dx}{dy} + \frac{1}{y}x = \frac{1}{y^{2}}\)
This is a linear differential equation in \(x\) of the form
\(\frac{dx}{dy} + P(y)x = Q(y)\)
Here, \(P(y)=\frac{1}{y}\) and \(Q(y)=\frac{1}{y^{2}}\)
The integrating factor (I.F.) is
\(\text{I.F.} = e^{\int \frac{1}{y}\,dy}\)
\(= e^{\log y}\)
\(= y\)
Multiplying the equation by I.F.,
\(y\frac{dx}{dy} + x = \frac{1}{y}\)
\(\Rightarrow \frac{d}{dy}(xy) = \frac{1}{y}\)
Integrating both sides,
\(xy = \int \frac{1}{y}dy + C\)
\(= \log y + C\)
\(\therefore \boxed{xy = \log y + C}\)
Solution:
Given,
\(2xy\frac{dy}{dx} = x^{2} + y^{2}\)
\(\Rightarrow \frac{dy}{dx} = \frac{x^{2}+y^{2}}{2xy}\)
\(\Rightarrow \frac{dy}{dx} = \frac{x}{2y} + \frac{y}{2x}\)
This is a homogeneous differential equation.
Let \(y = vx\)
\(\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\)
Substituting in the equation,
\(v + x\frac{dv}{dx} = \frac{1}{2v} + \frac{v}{2}\)
\(\Rightarrow 2v + 2x\frac{dv}{dx} = \frac{1}{v} + v\)
\(\Rightarrow 2x\frac{dv}{dx} = \frac{1}{v} - v\)
\(\Rightarrow 2x\frac{dv}{dx} = \frac{1-v^{2}}{v}\)
\(\Rightarrow \frac{2v}{1-v^{2}}\,dv = \frac{dx}{x}\)
Integrating both sides,
\(\int \frac{2v}{1-v^{2}}\,dv = \int \frac{dx}{x} - \log \, C \)
\(\Rightarrow -\log|1-v^{2}| = \log|x| - \log \, C\)
\(\Rightarrow \log|1-v^{2}| = -\log|x| + \log \, C\)
\(\Rightarrow 1-v^{2} = \frac{C}{x}\)
Substituting \(v=\frac{y}{x}\),
\(1 - \frac{y^{2}}{x^{2}} = \frac{C}{x}\)
\(\Rightarrow \frac{x^{2}-y^{2}}{x^{2}} = \frac{C}{x}\)
\(\Rightarrow x^{2}-y^{2} = Cx\)
\(\therefore \boxed{x^{2}-y^{2} = Cx}\)
Solution:
Given,
\(y\log y\,dx - x\,dy = 0\)
\(\Rightarrow y\log y\,dx = x\,dy\)
\(\Rightarrow \frac{dy}{dx} = \frac{y\log y}{x}\)
\(\Rightarrow \frac{1}{y\log y}\,dy = \frac{1}{x}\,dx\)
Integrating both sides,
\(\int \frac{1}{y\log y}\,dy = \int \frac{1}{x}\,dx + \log C \quad .... (i)\)
Let \(t=\log y\)
\(\Rightarrow dt=\frac{1}{y}dy\)
\(\Rightarrow \int \frac{1}{y\log y}\,dy = \int \frac{1}{t}\,dt\)
\(= \log|t|\)
\(= \log|\log y|\)
Therefore, equation \((i)\) becomes -
\(\Rightarrow \log|\log y| = \log|x| + \log C\)
\(\Rightarrow \log y = Cx\)
\(\Rightarrow y = e^{Cx}\)
\(\therefore \boxed{\log y = Cx \text{ or } y=e^{Cx}}\)
Hi Please, do not Spam in Comments.