Bobbie Comstock

2021-12-28

Solve the following differential equations: $y-{y}^{\prime }=5{e}^{x}-\mathrm{sin}\left(2x\right)$

deginasiba

$y-{y}^{\prime }=5{e}^{x}-\mathrm{sin}\left(2x\right)$
Auxilary equation is ${m}^{2}-m=0$
$⇒m\left(m-1\right)=0⇒m=0,1$
Complementary function $={c}_{1}+{c}_{2}{e}^{x}$
Particular solution $=\frac{1}{D\left(D-1\right)}5{e}^{x}-\frac{1}{{D}^{2}-D}\mathrm{sin}2x$
$=5\left(\frac{1}{D-1}-\frac{1}{D}\right){e}^{x}-\frac{1}{-4-D}\mathrm{sin}2x$
$=5\left(x{e}^{x}-{e}^{x}\right)+\frac{\left(D-4\right)}{\left(D+4\right)\left(D-4\right)}\mathrm{sin}2x$
$=5\left(x{e}^{x}-{e}^{x}\right)+\frac{\left(D-4\right)}{{D}^{2}-16}\mathrm{sin}2x$
$=5{e}^{x}\left(x-1\right)+\frac{\left(D-4\right)}{-4-16}\mathrm{sin}2x$
$=5{e}^{x}\left(x-1\right)-\frac{1}{20}\left(2\mathrm{cos}2x-4\mathrm{sin}2x\right)$
$y\left(x\right)={c}_{1}+{c}_{2}{e}^{x}+5{e}^{x}\left(x-1\right)-\frac{1}{10}\mathrm{cos}2x+\frac{1}{5}\mathrm{sin}2x$

Maricela Alarcon

Step 1
Given differential equation,
$y-{y}^{\prime }=5{e}^{x}-\mathrm{sin}\left(2x\right)$
Step 2
Evaluate complementary function:
${m}^{2}-m=0$
$m\left(m-1\right)=0$
$m=0,1$
Since roots are real and non repeated so complementary function is
${y}_{c}={c}_{1}{e}^{0x}+{c}_{2}{e}^{x}$
${y}_{c}={c}_{1}+{c}_{2}{e}^{x}$
Step 3
Now evaluate particular integral:
${y}_{p}=\frac{1}{{D}^{2}-D}\left(5{e}^{x}-\mathrm{sin}2x\right)$
$=\frac{1}{{D}^{2}-D}\left(5{e}^{x}\right)-\frac{1}{{D}^{2}-D}\left(\mathrm{sin}2x\right)$
$=5\frac{1}{{D}^{2}-D}\left({e}^{x}\right)-\frac{1}{{D}^{2}-D}\left(\mathrm{sin}2x\right)$
$=5\frac{x}{2D-1}{e}^{x}-\frac{1}{-4-D}\left(\mathrm{sin}2x\right)$
$=5\frac{x}{2\left(1\right)-1}{e}^{x}+\frac{1}{\left(D+4\right)}×\frac{\left(D-4\right)}{\left(D-4\right)}\left(\mathrm{sin}2x\right)$
$=5x{e}^{x}+\frac{\left(D-4\right)}{\left({D}^{2}-16\right)}\left(\mathrm{sin}2x\right)$
$=5x{e}^{x}+\frac{\left(D-4\right)}{\left(-4-16\right)}\left(\mathrm{sin}2x\right)$
$=5x{e}^{x}-\frac{1}{20}\left(D-4\right)\left(\mathrm{sin}2x\right)$
$=5x{e}^{x}-\frac{1}{20}\left(2\mathrm{cos}2x-4\mathrm{sin}2x\right)$
Step 4
Since complete solution of differential equation is sum of complementary function and particular integral, so
$y={y}_{c}+{y}_{p}$

karton

$\begin{array}{}{y}^{″}-{y}^{\prime }=5{e}^{x}-\mathrm{sin}\left(2x\right)\\ {m}^{2}-m=0\\ m\left(m-1\right)=0\\ m=0,1\\ {y}_{c}={c}_{1}{e}^{0x}+{c}_{2}{e}^{x}\\ {y}_{c}={c}_{1}+{c}_{2}{e}^{x}\\ {y}_{p}=\frac{1}{{D}^{2}-D}\left(5{e}^{x}-\mathrm{sin}2x\right)\\ =\frac{1}{{D}^{2}-D}\left(5{e}^{x}\right)-\frac{1}{{D}^{2}-D}\left(\mathrm{sin}2x\right)\\ =5\frac{1}{{D}^{2}-D}\left({e}^{x}\right)-\frac{1}{{D}^{2}-D}\left(\mathrm{sin}2x\right)\\ =5\frac{x}{2D-1}{e}^{x}-\frac{1}{-4-D}\left(\mathrm{sin}2x\right)\\ =5\frac{x}{2\left(1\right)-1}{e}^{x}+\frac{1}{\left(D+4\right)}×\frac{\left(D-4\right)}{\left(D-4\right)}\left(\mathrm{sin}2x\right)\\ =5x{e}^{x}+\frac{\left(D-4\right)}{\left({D}^{2}-16\right)}\left(\mathrm{sin}2x\right)\\ =5x{e}^{x}+\frac{\left(D-4\right)}{\left(-4-16\right)}\left(\mathrm{sin}2x\right)\\ =5x{e}^{x}-\frac{1}{20}\left(D-4\right)\left(\mathrm{sin}2x\right)\\ =5x{e}^{x}-\frac{1}{20}\left(2\mathrm{cos}2x-4\mathrm{sin}2x\right)\\ y={y}_{c}+{y}_{p}\\ y={c}_{1}+{c}_{2}{e}^{x}+5x{e}^{x}-\frac{1}{20}\left(2\mathrm{cos}2x-4\mathrm{sin}2x\right)\\ y={c}_{1}+{c}_{2}{e}^{x}+5x{e}^{x}-\frac{1}{20}\left(2\mathrm{cos}2x-4\mathrm{sin}2x\right)\end{array}$