Question

(1 + e^{2x})y^2 dy = e^x dx

Explanation:

Step1: Recognize the differential equation type

This is a separable differential equation, so we can integrate both sides separately. The equation is \((1 + e^{2x})y^{2}dy = e^{x}dx\).

Step2: Integrate the left - hand side

Integrate \(\int y^{2}dy\) with respect to \(y\). Using the power rule for integration \(\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 2\), we have \(\int y^{2}dy=\frac{y^{3}}{3}+C_1\). But we also have the factor \((1 + e^{2x})\) which is treated as a constant with respect to \(y\) during the integration with respect to \(y\). So the integral of the left - hand side \(\int(1 + e^{2x})y^{2}dy=(1 + e^{2x})\int y^{2}dy=(1 + e^{2x})\frac{y^{3}}{3}+C_1\).

Step3: Integrate the right - hand side

Integrate \(\int e^{x}dx\) with respect to \(x\). We know that the integral of \(e^{x}\) with respect to \(x\) is \(e^{x}+C_2\), so \(\int e^{x}dx = e^{x}+C_2\).

Step4: Combine the two results

Putting the two integrals together, we have \((1 + e^{2x})\frac{y^{3}}{3}=e^{x}+C\) (where \(C = C_2 - C_1\) is the combined constant of integration). We can also multiply both sides by 3 to get \((1 + e^{2x})y^{3}=3e^{x}+C'\) (where \(C'=3C\)). If we want to solve for \(y\), we can write \(y^{3}=\frac{3e^{x}+C'}{1 + e^{2x}}\), and then \(y=\sqrt[3]{\frac{3e^{x}+C'}{1 + e^{2x}}}\) or simplify the constant as needed.

Answer:

The general solution of the differential equation \((1 + e^{2x})y^{2}dy = e^{x}dx\) is \(\frac{(1 + e^{2x})y^{3}}{3}=e^{x}+C\) (or equivalent forms after simplifying the constant and solving for \(y\) further).