The solution of the differential equation sqr | vedclass

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

Question

(A) Given the differential equation $\sqrt{a + x} \frac{dy}{dx} + xy = 0$.Rearranging the terms,we get $\frac{dy}{dx} = -\frac{xy}{\sqrt{a + x}}$.Sepa

Vedclass · Accepted Answer

$y = A e^{\frac{2}{3}(2a - x)\sqrt{x + a}}$

Vedclass · Answer

(A) Given the differential equation $\sqrt{a + x} \frac{dy}{dx} + xy = 0$.Rearranging the terms,we get $\frac{dy}{dx} = -\frac{xy}{\sqrt{a + x}}$.Separating the variables,we have $\frac{dy}{y} = -\frac{x}{\sqrt{a + x}} dx$.Integrating both sides,$\int \frac{dy}{y} = -\int \frac{x}{\sqrt{x + a}} dx$.Let $u = x + a$,then $x = u - a$ and $dx = du$.$\ln y = -\int \frac{u - a}{\sqrt{u}} du = -\int (u^{1/2} - a u^{-1/2}) du$.$\ln y = -(\frac{2}{3} u^{3/2} - 2a u^{1/2}) + C$.$\ln y = -\frac{2}{3} (x + a)^{3/2} + 2a(x + a)^{1/2} + \ln A$.$\ln y = -\frac{2}{3} (x + a)^{1/2} (x + a) + 2a(x + a)^{1/2} + \ln A$.$\ln y = \sqrt{x + a} [-\frac{2}{3}x - \frac{2}{3}a + 2a] + \ln A$.$\ln y = \sqrt{x + a} [\frac{-2x - 2a + 6a}{3}] + \ln A = \frac{2}{3} \sqrt{x + a} (2a - x) + \ln A$.Taking the exponential of both sides,$y = A e^{\frac{2}{3}(2a - x)\sqrt{x + a}}$.

The solution of the differential equation $\sqrt{a + x} \frac{dy}{dx} + xy = 0$ is

Similar Questions

Let $y=f(x)$ be the solution of the differential equation $y(x+1) dx - x^2 dy = 0$ with the initial condition $y(1)=e$. Then $\lim _{x \rightarrow 0^{+}} f(x)$ is equal to

The solution curve of the differential equation,$(1+e^{-x})(1+y^{2}) \frac{dy}{dx} = y^{2}$,which passes through the point $(0,1)$,is

The solution of the differential equation $e^{-x}(y + 1)dy + (\cos^2 x - \sin 2x)y dx = 0$, given that $y = 1$ when $x = 0$ is

The general solution of the differential equation $\cos (x+y) \frac{dy}{dx} = 1$ is

The solution of the equation $\frac{dy}{dx} = y(e^x + 1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module