Question

Solve the initial-value problem y'+(2/x)y=-2xy/(x^2+2x^2*y+1), y(1)=-2.

Answer 1

@hartnn @amistre64

Answer 2

\[\begin{align*}
y'+\frac{2}{x}y&=-\frac{2xy}{x^2+2x^2y+1}\\\\
y'&=-\frac{4x^2y+4x^2y^2+2y}{x(x^2+2x^2y+1)}\\\\
4x^2y+4x^2y^2+2y+x(x^2+2x^2y+1)y'&=0
\end{align*}\]
Check for exactness:
\[\frac{\partial }{\partial y}\left[4x^2y+4x^2y^2+2y\right]=4x^2+8x^2y+2\\
\frac{\partial }{\partial x}\left[x^3+2x^3y+x\right]=3x^2+6x^2y+1\]
Integrating factor:
\[\begin{align*}\mu(x)&=\exp\left(\int\frac{(4x^2+8x^2y+2)-(3x^2+6x^2y+1)}{x^3+2x^3y+x}~dx\right)\\\\
&=\exp\left(\int\frac{x^2+2x^2y+1}{x^3+2x^3y+x}~dx\right)\\\\
&=\exp\left(\int\frac{dx}{x}\right)\\\\
&=x
\end{align*}\]
Distribute the IF:
\[\begin{align*}
4x^3y+4x^3y^2+2xy+(x^4+2x^4y+x^2)y'&=0
\end{align*}\]
Check for exactness again:
\[\frac{\partial }{\partial y}\left[4x^3y+4x^3y^2+2xy\right]=4x^3+8x^3y+2x\\
\frac{\partial }{\partial x}\left[x^4+2x^4y+x^2\right]=4x^3+8x^3y+2x\]
Now you're ready to solve for \(\Psi(x,y)\).
\[\begin{align*}\Psi_x&=4x^3y+4x^3y^2+2xy\\\\
\Psi&=\int\cdots~dx\\\\
&=x^4y+x^4y^2+x^2y+f(y)\\\\
\Psi_y&=x^4+2x^4y+x^2+f'(y)\\\\
x^4+2x^4y+x^2&=\\\\
f'(y)&=0\\\\
f(y)=C\end{align*}\]

Answer 3

Thank you!

Answer 4

You're welcome! I'm assuming the exact approach is fine by you? I was looking for a proper substitution the first time around, but I saw you posting questions about exact equations lately so I figured you wouldn't have trouble following this one.