Find the particular solution that satisfies the given initial condition
$(y^2+y)e^x{y}^{\prime }=y^3+e^x{y}^3;y\left(0\right)=1$

• A. $\ln \mid y\mid -\frac{1}{y}=-e^{-x}+x$
• B. $\ln \mid y\mid -\frac{1}{y}=-e^{-x}+x+1$
• C. $\ln \mid \frac{y}{y^2}\mid =-e^{-x}+x$
• D. $\ln \mid \frac{y}{y^2}\mid =-e^{-x}+x+1$

Answer: (A)

Verified step by step guidance

1. Step 1: Start by rewriting the given differential equation \((y^2 + y)e^x y' = y^3 + e^x y^3\). Factorize the terms on the right-hand side to simplify the equation. This gives \((y^2 + y)e^x y' = y^3(1 + e^x)\).
2. Step 2: Divide through by \((y^2 + y)\) to isolate \(y'\). This results in \(e^x y' = \frac{y^3(1 + e^x)}{y^2 + y}\). Simplify the fraction on the right-hand side by factoring \(y\) from the denominator, giving \(e^x y' = \frac{y^2(1 + e^x)}{y + 1}\).
3. Step 3: Rearrange the equation to separate variables. Divide both sides by \(e^x\) and multiply through by \(\frac{y + 1}{y^2}\), resulting in \(\frac{y + 1}{y^2} dy = (1 + e^x)e^{-x} dx\). Simplify the right-hand side to \(\frac{y + 1}{y^2} dy = (1 + e^{-x}) dx\).
4. Step 4: Integrate both sides. For the left-hand side, split \(\frac{y + 1}{y^2}\) into \(\frac{1}{y} + \frac{1}{y^2}\) and integrate term by term. For the right-hand side, integrate \(1 + e^{-x}\) with respect to \(x\). This gives \(\int \frac{1}{y} dy + \int \frac{1}{y^2} dy = \int 1 dx + \int e^{-x} dx\).
5. Step 5: Solve the integrals. The left-hand side becomes \(\ln|y| - \frac{1}{y}\), and the right-hand side becomes \(x - e^{-x} + C\), where \(C\) is the constant of integration. Use the initial condition \(y(0) = 1\) to solve for \(C\). Substitute \(C\) back into the equation to find the particular solution.