91Ó°ÊÓ

The differential equation is,

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}......\left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2-3\mathrm{m}+2=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-3\mathrm{m}+2=0\\ {\mathrm{m}}^2-2\mathrm{m}-\mathrm{m}+2=0\\ \mathrm{m}(\mathrm{m}-2)-1(\mathrm{m}-2)=0\\ (\mathrm{m}-1)(\mathrm{m}-2)=0\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=1,\text{ â¶Ä‰â¶Ä‰}\&\text{ â¶Ä‰â¶Ä‰}{\mathrm{m}}_2=2$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{x}}$

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

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}\\ \text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰}\mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}\end{array}\right\}\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰}…\left(2\right)$

For $\mathrm{β}≠0$, use the form;

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{³¦´Ç²õβ³Ù}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\mathrm{²õ¾\pm ²Ôβ³Ù}]$

With s = 1 if$\mathbf{Î\pm }\mathbf{+}\mathbf{¾\pm β}$ is a root of the associated auxiliary equation.

And s = 0 if$\mathbf{Î\pm }\mathbf{+}\mathbf{¾\pm β}$ is not a root of the associated auxiliary equation.

Comparing equations (1) and (2), we get:

M=0 and$\mathrm{Î\pm }=1;\mathrm{β}=1$

Therefore,$\mathrm{Î\pm }+\mathrm{¾\pm β}=1+\mathrm{i}$ is not a root of the associated auxiliary equation so here s = 0.

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{Î\pm ³\ae }}\mathrm{³¦´Ç²õβ³\ae }+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{Î\pm ³\ae }}\mathrm{²õ¾\pm ²Ôβ³\ae }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^0\left({\mathrm{A}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{x}}\cos \left(1\right)\mathrm{x}+{\mathrm{t}}^0\left({\mathrm{B}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{x}}\sin \left(1\right)\mathrm{x}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_0{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+{\mathrm{B}}_0{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰}…\left(3\right)\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})+{\mathrm{e}}^{\mathrm{x}}(-\mathrm{Asinx}+\mathrm{Bcosx})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=(-2\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\left(2\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),\text{ â¶Ä‰}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$the equation (1),

$\begin{array}{r}⇒\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ ⇒-\left(2\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\left(2\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-3[(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}]+2\left[{\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\right]\\ \text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ ⇒(\mathrm{A}-\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(-\mathrm{A}-\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}\mathrm{A}-\mathrm{B}=1\text{ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}......\left(4\right)\\ -\mathrm{A}-\mathrm{B}=0\end{array}$

Solve the above equations,

role="math" localid="1654943684164" $\begin{array}{l}\mathrm{A}-\mathrm{B}=1\\ -\mathrm{A}-\mathrm{B}=0\\ \stackrel{¯}{\begin{array}{l}\text{ â¶Ä‰}-2\mathrm{B}=1\text{ â¶Ä‰}\\ \text{​ â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰}\mathrm{B}=\frac{-1}{2}\end{array}}\end{array}$

Substitute the value of B in the equation (4),

role="math" localid="1654943962777" $\begin{array}{c}\mathrm{A}-\left(\frac{-1}{2}\right)=1\\ \mathrm{A}=\frac{1}{2}\end{array}$

Substitute the value of A and B in the equation (3),

role="math" localid="1654944097396" $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}\mathrm{cosx}-\frac{1}{2}\mathrm{sinx}\right)\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}\mathrm{cosx}-\frac{1}{2}\mathrm{sinx}\right)$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{x}}+\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}(\mathrm{cosx}-\mathrm{sinx})\end{array}$