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: $(x-2y+1)dy - (3x-6y+2)dx = 0$ $\Rightarrow \frac{dy}{dx} = \frac{3x-6y+2}{x-2y+1} = \frac{3(x-2y)+2}{(x-2y)+1}$ Let

Vedclass · Accepted Answer

$\left|x-2y+\frac{3}{5}\right|^{2/25} \cdot e^{1/5(6x-2y)}=C$

Vedclass · Answer

(C) Given differential equation: $(x-2y+1)dy - (3x-6y+2)dx = 0$ $\Rightarrow \frac{dy}{dx} = \frac{3x-6y+2}{x-2y+1} = \frac{3(x-2y)+2}{(x-2y)+1}$ Let $v = x-2y$. Then $\frac{dv}{dx} = 1 - 2\frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dv}{dx})$. Substituting this into the equation: $\frac{1}{2}(1 - \frac{dv}{dx}) = \frac{3v+2}{v+1}$ $1 - \frac{dv}{dx} = \frac{6v+4}{v+1} \Rightarrow \frac{dv}{dx} = 1 - \frac{6v+4}{v+1} = \frac{v+1-6v-4}{v+1} = \frac{-5v-3}{v+1}$ $\int \frac{v+1}{-5v-3} dv = \int dx$ $-\frac{1}{5} \int \frac{5v+5}{5v+3} dv = x + C_1$ $-\frac{1}{5} \int \frac{(5v+3)+2}{5v+3} dv = x + C_1$ $-\frac{1}{5} [v + \frac{2}{5} \ln|5v+3|] = x + C_1$ Substitute $v = x-2y$: $-\frac{1}{5}(x-2y) - \frac{2}{25} \ln|5(x-2y)+3| = x + C_1$ $-\frac{2}{25} \ln|5x-10y+3| = x + \frac{1}{5}x - \frac{2}{5}y + C_1 = \frac{6}{5}x - \frac{2}{5}y + C_1$ $\ln|5x-10y+3|^{-2/25} = \frac{6x-2y}{5} + C_1$ $|5x-10y+3|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot e^{C_1}$ $|5(x-2y+\frac{3}{5})|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_2$ $|x-2y+\frac{3}{5}|^{-2/25} = e^{\frac{6x-2y}{5}} \cdot C_3$ $\left|x-2y+\frac{3}{5}\right|^{2/25} \cdot e^{\frac{1}{5}(6x-2y)} = C$

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

Similar Questions

Let $y=y(x)$ be a solution curve of the differential equation,$(1-x^2 y^2) dx = y dx + x dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$,then a value of $\alpha$ is

General solution of differential equation $\frac{dy}{dx} + y = 1$ $(y \neq 1)$ is

$A$ differentiable function satisfies $3f^2(x) f'(x) = 2x$. Given $f(2) = 1$,then the value of $f(3)$ is:

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$, then $|f(xy)|$ is equal to

For solving $\frac{dy}{dx} = (4x + y + 1)$,the suitable substitution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module