Question

solve the given equation by variation of parameters.
y + 7y = 3x
the solution is y = c_1 + c_2 e^{7x} - \frac{3}{14}x^2 + \frac{3}{343}(7x - 1)

Explanation:

Step1: Find the complementary function

The homogeneous equation is \(y'' + 7y'=0\). The characteristic equation is \(r^{2}+7r = 0\), factoring gives \(r(r + 7)=0\). So \(r_1=0\) and \(r_2=-7\). The complementary function \(y_c=c_1 + c_2e^{-7x}\).

Step2: Set up for variation of parameters

We assume a particular solution \(y_p=u_1(x)+u_2(x)e^{-7x}\). The Wronskian \(W=

$$\begin{vmatrix}1&e^{-7x}\\0&-7e^{-7x}\end{vmatrix}$$

=-7e^{-7x}\).

For \(f(x) = 3x\), we calculate \(u_1'\) and \(u_2'\) using the formulas \(u_1'=\frac{

$$\begin{vmatrix}0&e^{-7x}\\3x&-7e^{-7x}\end{vmatrix}$$

}{W}\) and \(u_2'=\frac{

$$\begin{vmatrix}1&0\\0&3x\end{vmatrix}$$

}{W}\).

\(u_1'=\frac{- 3xe^{-7x}}{-7e^{-7x}}=\frac{3x}{7}\), integrating \(u_1=\frac{3}{14}x^{2}\).

\(u_2'=\frac{3x}{-7e^{-7x}}\times\frac{1}{e^{-7x}}=-\frac{3x}{7}e^{7x}\). Using integration by parts (\(\int xe^{ax}dx=\frac{xe^{ax}}{a}-\frac{e^{ax}}{a^{2}}+C\), here \(a = 7\)), \(u_2=-\frac{3}{49}(7x - 1)e^{7x}\).

Then \(y_p=u_1+u_2e^{-7x}=\frac{3}{14}x^{2}-\frac{3}{49}(7x - 1)\).

The general solution \(y=y_c + y_p=c_1 + c_2e^{-7x}-\frac{3}{14}x^{2}+\frac{3}{343}(7x - 1)\)

Answer:

\(y = c_1 + c_2e^{-7x}-\frac{3}{14}x^{2}+\frac{3}{343}(7x - 1)\)