Question

solve using integrating factor (x^5+3y)dx-xdy=0
tnx :D

Answer 1

Using Integrating factor\[x.\frac{dy}{dx}=x ^{5} + 3y => \frac{dy}{dx} - \frac{3}{x}y = x ^{4}\]

Now the Integrating factor is \[\exp [\int\limits_{}^{}\frac{-3}{x}dx] = \exp[-3lnx] = (\exp[lnx])^{-3} = x ^{-3}\]

Hence multiplying by \[x ^{-3}\]

We get \[\frac{1}{x ^{3}}.\frac{dy}{dx} - \frac{3}{x ^{4}}y = x\]
\[=> \frac{d}{dx}(\frac{y}{x^{3}}) = x\]
\[=> \frac{y}{x^{3}} = \frac{x^{2}}{2} + k\]
\[=> y = \frac{x^{5}}{2} + kx^{2}\]

Answer 2

Please let me know if you get it