The general solution of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$Divide by $(1-x^{2})$ to get the standard form $\frac{d

Vedclass · Accepted Answer

$y = \sqrt{1-x^{2}} + c(1-x^{2})$

Vedclass · Answer

(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$Divide by $(1-x^{2})$ to get the standard form $\frac{dy}{dx} + Py = Q$:$\frac{dy}{dx} + \frac{2x}{1-x^{2}}y = \frac{x}{\sqrt{1-x^{2}}}$Here,$P = \frac{2x}{1-x^{2}}$ and $Q = \frac{x}{\sqrt{1-x^{2}}}$.Integrating Factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1-x^{2}} dx} = e^{-\ln(1-x^{2})} = e^{\ln(\frac{1}{1-x^{2}})} = \frac{1}{1-x^{2}}$.The general solution is $y 	imes (I.F.) = \int Q 	imes (I.F.) dx + c$.$y 	imes \frac{1}{1-x^{2}} = \int \frac{x}{\sqrt{1-x^{2}}} 	imes \frac{1}{1-x^{2}} dx + c$$y 	imes \frac{1}{1-x^{2}} = \int x(1-x^{2})^{-\frac{3}{2}} dx + c$Let $u = 1-x^{2}$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.$y 	imes \frac{1}{1-x^{2}} = -\frac{1}{2} \int u^{-\frac{3}{2}} du + c = -\frac{1}{2} \left( \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} ight) + c = u^{-\frac{1}{2}} + c = \frac{1}{\sqrt{1-x^{2}}} + c$.Multiplying both sides by $(1-x^{2})$:$y = \frac{1-x^{2}}{\sqrt{1-x^{2}}} + c(1-x^{2}) = \sqrt{1-x^{2}} + c(1-x^{2})$.

The general solution of the differential equation $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$ is

Similar Questions

If $\cos x \frac{dy}{dx} - y \sin x = 6x$ for $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{6})$ is equal to:

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

If $\frac{dy}{dx} + \frac{y}{x} = x^2$,then $2y(2) - y(1) =$

Integrating factor of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is

The solution of the differential equation $x^2 \frac{dy}{dx} - xy = 1$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module