Solution

Given: \(y\,dx-(x+3y^{2})\,dy=0 \;\Rightarrow\; y\,\dfrac{dx}{dy}=x+3y^{2}\).

So \( \dfrac{dx}{dy}-\dfrac{1}{y}x=3y\) (linear). Integrating factor \(=\exp\!\int\!-\dfrac{1}{y}dy=\dfrac{1}{y}\).

\(\displaystyle \frac{d}{dy}\!\left(\frac{x}{y}\right)=3 \;\Rightarrow\; \frac{x}{y}=3y+C \;\Rightarrow\; x=3y^{2}+Cy.\)

Through \((1,1)\): \(1=3(1)+C(1)\Rightarrow C=-2\). Hence curve: \(x=3y^{2}-2y\).

Solution

Solution

$\dfrac{dy}{dx}=e^y(e^x+x^2)\ \Rightarrow\ e^{-y}dy=(e^x+x^2)dx$

$\int e^{-y}dy=\int (e^x+x^2)dx$ $\Rightarrow\ -e^{-y}=e^x+\dfrac{x^3}{3}+C$

Hence, $e^{-y}+e^x+\dfrac{x^3}{3}=C$ (or $y=-\ln\!\big(C-e^x-\dfrac{x^3}{3}\big)$).

Answer: $\boxed{e^{-y}+e^x+\dfrac{x^3}{3}=C}$ ✅