Question

In Problems 13–18 proceed as in Example 3 to find a solution of the

given initial-value problem. Evaluate the integral that defines \({y_p}\left( x \right)\).

15. \(y'' - 10y' + 25y = {e^{5x}},y(0) = 0,y'(0) = 0\)

Short answer

\({y_p}(x) = \frac{1}{2}{x^2}{e^{5x}}\)

Step-by-step solution

The second order differential equation

We have the second order differential equation

with the initial conditions

\(y(0) = 0 and {y^'}(0) = 0\)

and we have to get its solution as the following technique :

First, we have to find the solutions of the corresponding homogeneous differential equation

by assuming that \(y = {e^{mx}}\) is a solution for this homogeneous differential equation, then follow then differentiate this assumption with respect to \(x\), then we have

\(\begin{aligned}{l}{y^\prime } = m{e^{mx}}\\{y^{\prime \prime }} = {m^2}{e^{mx}}\end{aligned}\)

After that, substitute with \(y = {e^{mx}}\) and equations \((2)\)and (3) into the differential equation (1), then we obtain

\(\begin{aligned}{l}{m^2}{e^{mx}} - 10m{e^{mx}} + 25{e^{mx}} = 0\\{e^{mx}}\left( {{m^2} - 10m + 25} \right) = 0\end{aligned}\)

Since \({e^{mx}}\)cannot be equal 0, the we have

\(\begin{aligned}{l}{m^2} - 10m + 25 = 0\\{(m - 5)^2} = 0\end{aligned}\)

Then the roots are

\({m_1} = 5{\rm{ and }}{m_2} = 5\)

Then we can obtain the solution of the corresponding homogeneous differential equation as

\({y_c} = {k_1}{e^{5x}} + {k_2}x{e^{5x}}\)

Second, now we have to find the particular solution \({y_p}(x)\) of the given differential equation as the following technique:

Since we have \({y_1} = {e^{5t}}\) and \({y_2} = t{e^{5t}}\), then we can obtain the wronskian \(W\left( {{y_1},{y_2}} \right)\)as

\(\begin{aligned}{c}W\left( {{e^{5t}},t{e^{5t}}} \right) = \left| {\begin{aligned}{*{20}{l}}{{y_1}}&{{y_2}}\\{y_1^\prime }&{y_2^\prime }\end{aligned}} \right|\\ = \left| {\begin{aligned}{*{20}{c}}{{e^{5t}}}&{t{e^{5t}}}\\{5{e^{5t}}}&{5t{e^{5t}} + {e^{5t}}}\end{aligned}} \right|\\ = \left( {{e^{5t}}} \right) \times \left( {5t{e^{5t}} + {e^{5t}}} \right) - \left( {t{e^{5t}}} \right) \times \left( {5{e^{5t}}} \right)\\ = 5t{e^{10t}} + {e^{10t}} - 5t{e^{10t}}\\ = {e^{10t}}\end{aligned}\)

After that, we have to find the Green's function for the given differential equation as

\(\begin{aligned}{c}G(x,t) = \frac{{{y_1}(t){y_2}(x) - {y_1}(x){y_2}(t)}}{{W(t)}}\\ = \frac{{{e^{5t}} \times x{e^{5x}} - {e^{5x}} \times t{e^{5t}}}}{{{e^{10t}}}}\\ = \frac{{{e^{5t}} \times x{e^{5x}} - {e^{5x}} \times t{e^{5t}}}}{{{e^{10t}}}} \times \frac{{{e^{ - 10t}}}}{{{e^{ - 10t}}}}\\ = \frac{{{e^{5t}} \times x{e^{5x}} \times {e^{ - 10t}} - {e^{5x}} \times t{e^{5t}} \times {e^{ - 10t}}}}{{{e^{10t}} \times {e^{ - 10x}}}}\\ = \frac{{{e^{ - 5t}} \times x{e^{5x}} - {e^{5x}} \times t{e^{ - 5t}}}}{1}\\ = x{e^{5x}}{e^{ - 5t}} - t{e^{ - 5t}}{e^{5x}}\end{aligned}\)

Using the Green’s function

Then, since \(f(t) = {e^{5t}}\)and using the Green's function we have to find the particular solution \({y_p}(x)\)as

\(\begin{aligned}{c}{y_p}(x) = \int_0^x G (x,t)f(t)dt\\ = \int_0^x {\left( {x{e^{5x}}{e^{ - 5t}} - t{e^{ - 5t}}{e^{5x}}} \right)} {e^{5t}}dt\\ = \int_0^x {\left( {x{e^{5x}} - t{e^{5x}}} \right)} dt\\ = x{e^{5x}}\int_0^x d t - {e^{5x}}\int_0^x t dt\\ = x{e^{5x}}(t)_0^x - {e^{5x}}\left( {\frac{1}{2}{t^2}} \right)_0^x\\ = x{e^{5x}}(x - 0) - \frac{1}{2}{e^{5x}}\left( {{x^2} - 0} \right)\\ = {x^2}{e^{5x}} - \frac{1}{2}{x^2}{e^{5x}}\\ = \frac{1}{2}{x^2}{e^{5x}}\end{aligned}\)

Final proof

\({y_p}(x) = \frac{1}{2}{x^2}{e^{5x}}\)