The general solution of the differential equa | vedclass

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

Question

(A) Given the differential equation $\frac{dy}{dx} = \frac{2x+y-3}{2y-x+3}$.Let $x = X+h$ and $y = Y+k$. Then $\frac{dy}{dx} = \frac{dY}{dX}$.We solve

Vedclass · Accepted Answer

$x^2 - xy - y^2 + 3x + 3y + c = 0$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = \frac{2x+y-3}{2y-x+3}$.Let $x = X+h$ and $y = Y+k$. Then $\frac{dy}{dx} = \frac{dY}{dX}$.We solve the system $2h+k-3=0$ and $-h+2k+3=0$.Multiplying the second by $2$: $-2h+4k+6=0$. Adding to the first: $5k+3=0 \implies k = -3/5$. Then $2h = 3 - (-3/5) = 18/5 \implies h = 9/5$.The equation becomes $\frac{dY}{dX} = \frac{2X+Y}{2Y-X}$.Let $Y = vX$,then $\frac{dY}{dX} = v + X\frac{dv}{dX}$.$v + X\frac{dv}{dX} = \frac{2+v}{2v-1} \implies X\frac{dv}{dX} = \frac{2+v-2v^2+v}{2v-1} = \frac{-2v^2+2v+2}{2v-1}$.Separating variables: $\int \frac{2v-1}{-2v^2+2v+2} dv = \int \frac{dX}{X}$.Let $u = -2v^2+2v+2$,then $du = (-4v+2) dv = -2(2v-1) dv$.So,$-\frac{1}{2} \int \frac{du}{u} = \ln|X| + C \implies -\frac{1}{2} \ln|-2v^2+2v+2| = \ln|X| + C$.$\ln|-2(Y/X)^2+2(Y/X)+2| = -2\ln|X| + C' \implies -2Y^2+2YX+2X^2 = C''$.Substituting $X=x-9/5$ and $Y=y+3/5$,we simplify to $x^2-xy-y^2+3x+3y+c=0$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x+y-3}{2y-x+3}$ is

Similar Questions

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Then the equation of the curve is:

The particular solution of the differential equation,$x y \frac{dy}{dx} = x^2 + 2y^2$ when $y(1) = 0$ is

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

If $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$, $x > 0$ and $y(1) = \frac{\pi}{2}$, then the value of $\cos \left(\frac{y}{x}\right)$ is

If $(x^2+y^2) dy = xy dx$,with $y(x_0) = e$ and $y(1) = 1$,then $x_0$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module