Question

Find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{11}\mathbf{t}\mathbf{+}\mathbf{1}$

Short answer

The general solution is$\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+11\mathrm{t}-1.$

Step-by-step solution

Write the auxiliary equation of the given differential equation

The differential equation is,

$\mathrm{y}\text{'}\text{'}-\mathrm{y}=-11\mathrm{t}+1\dots \left(1\right)$

The auxiliary equation for the above equation,

${\mathrm{m}}^2-1=0$

Now find the complementary solution of the given equation is

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-1=0\\ \mathrm{m}=\pm 1\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=1,\&{\mathrm{m}}_2=-1$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}$

Find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{At}+\mathrm{B}\dots \left(2\right)$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=\mathrm{A}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=0\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)$and${\mathbf{y}}_p"\left(\mathbf{x}\right)$the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}-\mathrm{y}=-11\mathrm{t}+1\\ 0-(\mathrm{At}+\mathrm{B})=-11\mathrm{t}+1\\ -\mathrm{At}-\mathrm{B}=-11\mathrm{t}+1\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}-\mathrm{A}=-11\Rightarrow \mathrm{A}=11\\ -\mathrm{B}=1\Rightarrow \mathrm{B}=-1\end{array}$

Substitute the value of A and B in the equation (2),

Therefore, the particular solution of equation (1),

$\begin{array}{l}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{At}+\mathrm{B}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=11\mathrm{t}-1\end{array}$

Conclusion.

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+11\mathrm{t}-1\end{array}$