Question

dy/dx=〖(-5x+y)〗^2

Answer 1

u = y -5x

u' = y' - 5 ---> y' = u' +5 = u^2

\[\rightarrow \frac{du}{dx} = u^2 -5\]
\[\int\limits \frac{du}{u^2 -5} = \int\limits dx\]
\[\frac{1}{2 \sqrt{5}}[ \ln(u-\sqrt{5}) - \ln (u+\sqrt{5}) = x + C\]
\[\ln \frac{u-\sqrt{5}}{u+\sqrt{5}} = 2 \sqrt{5} x +C\]
\[\frac{u-\sqrt{5}}{u+\sqrt{5}} = C e^{2 \sqrt{5} x}\]
\[u = \frac{\sqrt{5}(1+Ce^{2 \sqrt{5} x})}{1 - Ce^{2 \sqrt{5}x}}\]
\[y = \frac{\sqrt{5}(1+Ce^{2 \sqrt{5} x})}{1 - Ce^{2 \sqrt{5}x}} +5x\]