The general solution of the differential equa | vedclass

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(B) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}$.Let $x = X+h$ and $y = Y+k$. Choosing $h$ and $k$ such that $h+2k-3=0$ and

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$(x+y-2) = c(x-y)^3$

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(B) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-3}{2x+y-3}$.Let $x = X+h$ and $y = Y+k$. Choosing $h$ and $k$ such that $h+2k-3=0$ and $2h+k-3=0$,we find $h=1$ and $k=1$.Substituting $x=X+1$ and $y=Y+1$,the equation becomes $\frac{dY}{dX} = \frac{X+2Y}{2X+Y}$.This is a homogeneous differential equation. Let $Y = vX$,then $\frac{dY}{dX} = v + X\frac{dv}{dX}$.So,$v + X\frac{dv}{dX} = \frac{1+2v}{2+v}$.$X\frac{dv}{dX} = \frac{1+2v-2v-v^2}{2+v} = \frac{1-v^2}{2+v}$.Separating variables: $\int \frac{v+2}{1-v^2} dv = \int \frac{dX}{X}$.$\int (\frac{v}{1-v^2} + \frac{2}{1-v^2}) dv = \ln|X| + C$.$-\frac{1}{2}\ln|1-v^2| + 2 \cdot \frac{1}{2} \ln|\frac{1+v}{1-v}| = \ln|X| + C$.$-\frac{1}{2}\ln|(1-v)(1+v)| + \ln|\frac{1+v}{1-v}| = \ln|X| + C$.$\ln|\frac{1+v}{1-v}| - \frac{1}{2}\ln|1+v| - \frac{1}{2}\ln|1-v| = \ln|X| + C$.$\frac{1}{2}\ln|1+v| - \frac{3}{2}\ln|1-v| = \ln|X| + C$.$\ln|\frac{(1+v)^{1/2}}{(1-v)^{3/2}}| = \ln|X| + C$.$\frac{1+v}{(1-v)^3} = c X^2$. Substituting $v = Y/X = (y-1)/(x-1)$ and $X=x-1$:$\frac{1 + \frac{y-1}{x-1}}{(1 - \frac{y-1}{x-1})^3} = c(x-1)^2 \implies \frac{x+y-2}{(x-y)^3} = c \implies (x+y-2) = c(x-y)^3$.

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

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