Question

Find a particular solution to the differential equation.

$\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{3}{\mathbf{t}}^2$

Short answer

The particular solution is${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^2-12\mathrm{t}+24.$

Step-by-step solution

Use the method of undetermined coefficients to find a particular solution to the differential equation.

Consider the given differential equation,

$2\mathrm{x}\text{'}+\mathrm{x}=3{\mathrm{t}}^2\dots \left(1\right)$

According to the method of undetermined coefficients, the particular solution of the differential equation;

$\mathrm{ax}\text{'}\text{'}+\mathrm{bx}\text{'}+\mathrm{cx}={\mathrm{dt}}^{\mathrm{m}},\mathrm{m}=0,1,2,3,...$

It is of the form${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+{\mathrm{A}}_{\mathrm{m}-1}{\mathrm{t}}^{\mathrm{m}-1}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0$

Comparing the above equation with equation (1),

We get, m = 2

Find a particular solution to the differential equation for m = 2

Therefore, the particular solution of equation (1),

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0\dots \left(2\right)$

Now find the derivative of above equation,

${\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2{\mathrm{A}}_2\mathrm{t}+{\mathrm{A}}_1$

From the equation (1), substitute the value of ${\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$and ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$, we get

$\begin{array}{c}2{\mathrm{x}}_{\mathrm{p}}\text{'}+{\mathrm{x}}_{\mathrm{p}}=3{\mathrm{t}}^2\\ 2[2{\mathrm{A}}_2\mathrm{t}+{\mathrm{A}}_1]+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0=3{\mathrm{t}}^2\\ 4{\mathrm{A}}_2\mathrm{t}+2{\mathrm{A}}_1+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0=3{\mathrm{t}}^2\\ {\mathrm{A}}_2{\mathrm{t}}^2+[4{\mathrm{A}}_2+{\mathrm{A}}_1]\mathrm{t}+2{\mathrm{A}}_1+{\mathrm{A}}_0=3{\mathrm{t}}^2\end{array}$

Final conclusion.

Comparing the all coefficients of the above equation,

$\begin{array}{c}{\mathrm{A}}_2=3\\ 4{\mathrm{A}}_2+{\mathrm{A}}_1=0\dots \left(3\right)\\ 2{\mathrm{A}}_1+{\mathrm{A}}_0=0\dots \left(4\right)\end{array}$

Substitute the value of ${\mathrm{A}}_2$ in the equation (3),

$\begin{array}{c}4\left(3\right)+{\mathrm{A}}_1=0\\ {\mathrm{A}}_1=-12\end{array}$

Substitute the value of ${\mathrm{A}}_1$ in the equation (4),

$\begin{array}{c}2(-12)+{\mathrm{A}}_0=0\\ {\mathrm{A}}_0=24\end{array}$

Substitute the value of ${\mathrm{A}}_0,{\mathrm{A}}_1$and${\mathrm{A}}_2$ in the equation (2),

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^2-12\mathrm{t}+24$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0\\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^2+(-12)\mathrm{t}+24\\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^2-12\mathrm{t}+24\end{array}$