Find the general solutions homework help: x(x+y)y′=y(x−y)
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We are give to solve the ODE:
\(\displaystyle \displaystyle x(x+y)y'=y(x-y)\)
As you were moving towards a homogeneous equation, let's divide through by \(\displaystyle \displaystyle x^2 \) to obtain:
\(\displaystyle \displaystyle \left(1+\frac{y}{x}\right)y'=\frac{y}{x}\left(1-\frac{y}{x}\right)\)
Now, letting:
\(\displaystyle \displaystyle v=\frac{y}{x}\implies y'=v+v'x\)
We obtain:
\(\displaystyle \displaystyle (1+v)(v+v'x)=v(1-v)\)
\(\displaystyle \displaystyle v+xv'+v^2+xvv'=v-v^2\)
\(\displaystyle \displaystyle \left(v^{-2}+v^{-1}\right)v'=-2x^{-1}\)
Integrate with respect to \(\displaystyle x\):
\(\displaystyle \displaystyle \int \left(v^{-2}+v^{-1}\right)v'\,dx=-2\int x^{-1}\,dx\)
Can you continue?
\(\displaystyle \displaystyle x(x+y)y'=y(x-y)\)
As you were moving towards a homogeneous equation, let's divide through by \(\displaystyle \displaystyle x^2 \) to obtain:
\(\displaystyle \displaystyle \left(1+\frac{y}{x}\right)y'=\frac{y}{x}\left(1-\frac{y}{x}\right)\)
Now, letting:
\(\displaystyle \displaystyle v=\frac{y}{x}\implies y'=v+v'x\)
We obtain:
\(\displaystyle \displaystyle (1+v)(v+v'x)=v(1-v)\)
\(\displaystyle \displaystyle v+xv'+v^2+xvv'=v-v^2\)
\(\displaystyle \displaystyle \left(v^{-2}+v^{-1}\right)v'=-2x^{-1}\)
Integrate with respect to \(\displaystyle x\):
\(\displaystyle \displaystyle \int \left(v^{-2}+v^{-1}\right)v'\,dx=-2\int x^{-1}\,dx\)
Can you continue?