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# True-or-False: The solution of the differential equation $\large\frac{dy}{dx} = \frac{x+2y}{x}$ is $x+y=kx^2$.

$\begin{array}{1 1} True \\ False \\ cannot be determined \end{array}$

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Toolbox:
• A homogenous linear differential equation can be solved by putting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large \frac{dv}{dx} • \int \large\frac{dx}{1+x}$$=\log |1+x|+c$
$\large\frac{dy}{dx}=\frac{x+2y}{x}$
This is a homogenous linear differential equation hence this can be solved by substituting $y=vx$ and $\large\frac{dy}{dx}$$=v+x \large \frac{dv}{dx} v+x \large \frac{dv}{dx}=\frac{x+2vx}{x} v+x \large \frac{dv}{dx}=\frac{x(1+2v)}{x} =>x \large \frac{dv}{dx}$$=1+2v-v$
$=>x \large \frac{dv}{dx}$$=1+v Now seperating the variables we get, \large\frac{dv}{1+v}=\frac{dx}{x} Now integrating on both sides we get, \large \int \frac{dv}{1+v}=\int \frac{dx}{x} \log |1+v|=\log x +\log k Substituting for v=y/x we get, \log |1+\large\frac{y}{x}|$$=\log kx$