The solution of frac{dx}{dy} + frac{x}{y} = x | vedclass

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

Question

(B) Given the differential equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$ Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the differential equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$ Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1$ Let $t = \frac{1}{x}$, then $\frac{dt}{dy} = -\frac{1}{x^2} \frac{dx}{dy}$ Substituting this into the equation: $-\frac{dt}{dy} + \frac{t}{y} = 1$ Rearranging gives: $\frac{dt}{dy} - \frac{t}{y} = -1$ This is a linear differential equation of the form $\frac{dt}{dy} + P(y)t = Q(y)$, where $P(y) = -\frac{1}{y}$ and $Q(y) = -1$ Integrating Factor ($I$.$F$.) $= e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$ The general solution is $t \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$ $t \cdot \frac{1}{y} = \int (-1) \cdot \frac{1}{y} dy + c$ $\frac{1}{xy} = -\log y + c$ Multiplying by $y$, we get: $\frac{1}{x} = cy - y \log y$

The solution of $\frac{dx}{dy} + \frac{x}{y} = x^2$ is:

Similar Questions

The differential equation $\frac{dy}{dx} = \frac{x+y}{1+x^2}$ is a . . . . . . differential equation.

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

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

$A$ function $f(x)$ satisfies the condition $f(x) = f'(x) + f''(x) + f'''(x) + \dots \infty$,where $f(x)$ is an indefinitely differentiable function and the dash denotes the order of the derivative. If $f(0) = 1$,then $f(x)$ is:

If $y = y(x)$,$x \in \left(0, \frac{\pi}{2}\right)$ is the solution curve of the differential equation $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 2 \sin 4x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$,with $y\left(\frac{\pi}{4}\right) = e^{-\pi}$,then $y\left(\frac{\pi}{6}\right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module