91Ó°ÊÓ

Let the given function be \(y = {c_1}{e^{2x}} + {c_2}x{e^{2x}}\).

Then, the first derivative of the function is,

\(\frac{{dy}}{{dx}} = 2{c_1}{e^{2x}} + {c_2}{e^{2x}} + 2{c_2}x{e^{2x}}\)

The second derivative of the function is,

\(\begin{aligned}{l}\frac{{{d^2}y}}{{d{x^2}}} &= 4{c_1}{e^{2x}} + 2{c_2}{e^{2x}} + 2{c_2}{e^{2x}} + 4{c_2}x{e^{2x}}\\\frac{{{d^2}y}}{{d{x^2}}} &= 4{c_1}{e^{2x}} + 4{c_2}{e^{2x}} + 4{c_2}x{e^{2x}}\end{aligned}\)

Substitute \(y\) and \(y'\) into the left-hand side of the differential equation.

\(\begin{aligned}{c}\left( {4{c_1}{e^{2x}} + 4{c_2}{e^{2x}} + 4{c_2}x{e^{2x}}} \right) - 4\left( {2{c_1}{e^{2x}} + {c_2}{e^{2x}} + 2{c_2}x{e^{2x}}} \right) + 4\left( {{c_1}{e^{2x}} + {c_2}x{e^{2x}}} \right) &= 0\\{e^{2x}}\left( {4{c_1} + 4{c_2} + 4{c_2}x - 8{c_1} - 4{c_2} - 8{c_2}x + 4{c_1} + 4{c_2}x} \right) &= 0\\0 &= 0\end{aligned}\)

That is same as the right-hand side of the differential equation for any real values of \(x\). Thus, the indicated function is a solution of the given differential equation.