Question

solve the given equation by variation of parameters.
y + 2y = 2x

the solution is y = c₁ + c₂eˣ - x² + 2(-x - 1)

Explanation:

Step1: Find the complementary function

The homogeneous equation is \(y'' + 2y'=0\). The characteristic equation is \(r^{2}+2r = 0\), factoring gives \(r(r + 2)=0\). So \(r_{1}=0\) and \(r_{2}=-2\). The complementary function \(y_{c}=c_{1}+c_{2}e^{-2x}\).

Step2: Use variation of parameters

We assume \(y_{p}=u_{1}+u_{2}e^{-2x}\). Then \(y_{p}'=u_{1}'+u_{2}'e^{-2x}-2u_{2}e^{-2x}\). Let \(u_{1}'+u_{2}'e^{-2x}=0\), so \(y_{p}'=-2u_{2}e^{-2x}\), and \(y_{p}''=-2u_{2}'e^{-2x}+4u_{2}e^{-2x}\).
Substitute \(y_{p}\), \(y_{p}'\) and \(y_{p}''\) into \(y'' + 2y'=2x\):
\(-2u_{2}'e^{-2x}+4u_{2}e^{-2x}+2(-2u_{2}e^{-2x})=2x\), simplifies to \(-2u_{2}'e^{-2x}=2x\), so \(u_{2}'=-xe^{2x}\).
Integrate \(u_{2}'\) using integration by parts (\(\int xe^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C\)), \(u_{2}=-\frac{1}{2}xe^{2x}+\frac{1}{4}e^{2x}\).
Since \(u_{1}'+u_{2}'e^{-2x}=0\), \(u_{1}'=xe^{2x}\cdot e^{-2x}=x\). Integrate \(u_{1}'\), \(u_{1}=\frac{1}{2}x^{2}\).
\(y_{p}=\frac{1}{2}x^{2}+(-\frac{1}{2}xe^{2x}+\frac{1}{4}e^{2x})e^{-2x}=\frac{1}{2}x^{2}-x - 1\)

Step3: Write the general solution

The general solution \(y=y_{c}+y_{p}=c_{1}+c_{2}e^{-2x}+\frac{1}{2}x^{2}-x - 1\) (Note: There might be a typo in the given answer, the correct characteristic roots lead to \(e^{-2x}\) instead of \(e^{x}\))

Answer:

\(y = c_{1}+c_{2}e^{-2x}-x^{2}/2 - x - 1\)