Question

solve the given equation by variation of parameters.
y + y = -2x
the solution is y = \square.

Explanation:

Step1: Find the complementary solution

The homogeneous equation is \(y'' + y'=0\). Let \(r\) be the root of the characteristic equation \(r^{2}+r = 0\), factoring gives \(r(r + 1)=0\). So \(r_1=0\) and \(r_2=-1\). The complementary solution \(y_c=c_1 + c_2e^{-x}\).

Step2: Set up for variation of parameters

Let \(y_1 = 1\) and \(y_2=e^{-x}\). The Wronskian \(W=

$$\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}$$

=

$$\begin{vmatrix}1&e^{-x}\\0&-e^{-x}\end{vmatrix}$$

=-e^{-x}\).

The particular solution \(y_p=-y_1\int\frac{y_2g(x)}{W}dx + y_2\int\frac{y_1g(x)}{W}dx\), where \(g(x)=- 2x\).

First, \(\int\frac{y_2g(x)}{W}dx=\int\frac{e^{-x}\cdot(-2x)}{-e^{-x}}dx=\int2xdx=x^{2}\).

Second, \(\int\frac{y_1g(x)}{W}dx=\int\frac{1\cdot(-2x)}{-e^{-x}}dx = 2\int xe^{x}dx\). Using integration by parts (\(u = x\), \(dv=e^{x}dx\)), then \(du = dx\), \(v = e^{x}\). So \(2\int xe^{x}dx=2(xe^{x}-e^{x})\).

Then \(y_p=-1\cdot x^{2}+e^{-x}\cdot2(xe^{x}-e^{x})=-x^{2}+2x - 2\).

Step3: Write the general solution

The general solution \(y=y_c + y_p=c_1 + c_2e^{-x}-x^{2}+2x - 2\).

Answer:

\(y = c_1 + c_2e^{-x}-x^{2}+2x - 2\)