The general solution of the differential equa | vedclass

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

Question

(C) Given differential equation is $\frac{dx}{dy} + \frac{x}{y} = x^2$. Dividing both sides by $x^2$,we get $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy

Vedclass · Accepted Answer

$\frac{1}{x} = cy - y \log y$

Vedclass · Answer

(C) Given differential equation is $\frac{dx}{dy} + \frac{x}{y} = x^2$. Dividing both sides by $x^2$,we get $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$. Let $t = \frac{1}{x}$,then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$. Substituting this into the equation,we get $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$. This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$. The integrating factor $I.F. = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$. The solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. $t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$. $\frac{t}{y} = -\log |y| + c$. Substituting $t = \frac{1}{x}$,we get $\frac{1}{xy} = -\log |y| + c$,or $\frac{1}{x} = cy - y \log |y|$.

The general solution of the differential equation $\frac{dx}{dy} + \frac{x}{y} = x^2$ is

Similar Questions

Let $y=y(x)$ be the solution curve of the differential equation $(1+x^{2})dy+(y-\tan^{-1}x)dx=0$,with $y(0)=1$. Then the value of $y(1)$ is:

The integrating factor of $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + (\tan x)y = \sin x, 0 \leq x \leq \frac{\pi}{3},$ with $y(0) = 0,$ then $y\left(\frac{\pi}{4}\right)$ is equal to:

The solution of $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$ is

Let $y=y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1-x) \tan x + x(2-x)\} \, dx = 0$ such that $y(0)=2$. Then $y(2)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module