The general solution of the differential equa | vedclass

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

Question

(B) Let $v = 9x + y + 5$. Then $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$.Substituting into the equation: $\frac{dv}{

Vedclass · Accepted Answer

$\frac{1}{3} 	an^{-1} (\frac{9x + y + 5}{3}) = x + c$

Vedclass · Answer

(B) Let $v = 9x + y + 5$. Then $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$.Substituting into the equation: $\frac{dv}{dx} - 9 = v^2 \implies \frac{dv}{dx} = v^2 + 9$.Separating variables: $\int \frac{dv}{v^2 + 3^2} = \int dx$.Using the integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$, we get $\frac{1}{3} 	an^{-1}(\frac{v}{3}) = x + c$.Substituting $v = 9x + y + 5$ back: $\frac{1}{3} 	an^{-1}(\frac{9x + y + 5}{3}) = x + c$.

The general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is...

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$ is

The equation of the curve passing through $(3, 9)$ which satisfies the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$ is

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

Let $y=y(x)$ be the solution of the differential equation $(1+y^2) e^{\tan x} dx + \cos^2 x(1+e^{2 \tan x}) dy = 0$,with $y(0)=1$. Then $y(\frac{\pi}{4})$ is equal to:

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module