quiquenobi2v6

2021-12-28

Find the general solution of differential equations.
$y{}^{″}+{y}^{\prime }-2y={e}^{x}+4\mathrm{sin}x+{x}^{2}-x$

Serita Dewitt

Step 1
The given differential equation is
$y{}^{″}+{y}^{\prime }-2y={e}^{x}+4\mathrm{sin}x+{x}^{2}-x$
To find the solution, CF and PI calculated
Step 2
The auxilary equation can be written as
${m}^{2}+m-2=0$
${m}^{2}+2m-m-2=0$
$m\left(m+2\right)-1\left(m+2\right)=0$
$\left(m+2\right)\left(m-1\right)=0$
$⇒m=-2,1$
Complementry function
$CF={C}_{1}{e}^{m,x}+{c}_{2}{e}^{m2x}$
$CF={C}_{1}{e}^{x}+{C}_{2}{e}^{-2x}$
Step 3
To find particular integration
$PI=\frac{i}{{D}^{2}+D-2}{e}^{x}+\frac{4}{{D}^{2}+D-2}\mathrm{sin}x+\frac{1}{{D}^{2}+D-2}\left({x}^{2}-x\right)$
$=\frac{x.i{e}^{x}}{2D+1}+\frac{4}{-{1}^{2}+D-2}\mathrm{sin}x+\frac{-1}{2\left[1-\frac{D}{2}-\frac{{D}^{2}}{2}\right]}\left({x}^{2}-x\right)$
$=\frac{x.{e}^{x}}{2×1+1}+\frac{4}{D-3}\mathrm{sin}x-\frac{1}{2}{\left(1-\frac{D}{2}-\frac{{D}^{2}}{2}\right)}^{-1}\left({x}^{2}-x\right)$
$=\frac{x.{e}^{x}}{3}+\frac{4\left(D+3\right)}{{D}^{2}-9}\mathrm{sin}x\frac{-1}{2}\left[1+\left(\frac{D}{2}+\frac{{D}^{2}}{2}\right)+\frac{\left(-1\right)\left(-2\right)}{2}{\left(\frac{D+{D}^{2}}{2}\right)}^{2}\right]\left[{x}^{2}-x\right]$

Heather Fulton

For the equation ${y}^{″}+{y}^{\prime }-2y={e}^{x}+4\mathrm{sin}x+{x}^{2}-x$, the characteristic polynomium is ${m}^{2}+m-2=\left(m-1\right)\left(m+2\right)=0$. Roots 1 , - 2. The solution to the homogeneous part of the equation is $yh=C1{e}^{x}+C2{e}^{-2x}$. For the paticular integral we use the method of undetermined coefficients. Try $yp=Ax{e}^{x}+B\mathrm{sin}x+C\mathrm{cos}x+D{x}^{2}+Ex+F$. Substituting in the equation and collecting terms obtain $A=\frac{1}{3},B=-\frac{6}{5},C=-\frac{2}{5},D=-\frac{1}{2},E=0,F=-\frac{1}{2}$. Then the solution is $y=C1{e}^{x}+C2{e}^{-2x}+\left(\frac{1}{3}\right)x{e}^{x}-\left(\frac{6}{5}\right)\mathrm{sin}x-\left(\frac{2}{5}\right)\mathrm{cos}x-\left(\frac{1}{2}\right)\left({x}^{2}+1\right)$.

karton

Solve $\left({d}^{2}y\left(x\right)\right)/\left(d{x}^{2}\right)+\left(dy\left(x\right)\right)/\left(dx\right)-2y\left(x\right)={e}^{x}+{x}^{2}-x+4\mathrm{sin}\left(x\right):$
Apply the Laplace transformation ${L}_{x}\left[f\left(x\right)\right]\left(s\right)={\text{integral}}_{0}^{\mathrm{\infty }}f\left(x\right){e}^{-sx}dx$ to both sides:
${L}_{x}\left[\left({d}^{2}y\left(x\right)\right)/\left(d{x}^{2}\right)+\left(dy\left(x\right)\right)/\left(dx\right)-2y\left(x\right)\right]\left(s\right)={L}_{x}\left[{e}^{x}+{x}^{2}-x+4\mathrm{sin}\left(x\right)\right]\left(s\right)$
Evaluate Laplace transforms:
$\left({s}^{2}+s-2\right)\left({L}_{x}\left[y\left(x\right)\right]\left(s\right)\right)-\left(s+1\right)y\left(0\right)-{y}^{\prime }\left(0\right)=-1/{s}^{2}+2/{s}^{3}+1/\left(s-1\right)+4/\left({s}^{2}+1\right)$
Solve for ${L}_{x}\left[y\left(x\right)\right]\left(s\right)$:
${L}_{x}\left[y\left(x\right)\right]\left(s\right)=\left(-1/{s}^{2}+2/{s}^{3}+1/\left(s-1\right)+4/\left({s}^{2}+1\right)+y\left(0\right)\left(s+1\right)+{y}^{\prime }\left(0\right)\right)/\left({s}^{2}+s-2\right)$
Decompose ${L}_{x}\left[y\left(x\right)\right]\left(s\right)$ via partial fractions:
${L}_{x}\left[y\left(x\right)\right]\left(s\right)=1/\left(3\left(s-1{\right)}^{2}\right)+8/\left(9\left(s-1\right)\right)-1/{s}^{3}-1/\left(2s\right)+1/\left(90\left(s+2\right)\right)-6/\left(5\left({s}^{2}+1\right)\right)-\left(2s\right)/\left(5\left({s}^{2}+1\right)\right)+\left(2y\left(0\right)\right)/\left(3\left(s-1\right)\right)+y\left(0\right)/\left(3\left(s+2\right)\right)+\left({y}^{\prime }\left(0\right)\right)/\left(3\left(s-1\right)\right)-\left({y}^{\prime }\left(0\right)\right)/\left(3\left(s+2\right)\right)$
Take the inverse Laplace transform of both sides:
$y\left(x\right)=\left(8{e}^{x}\right)/9+{e}^{-2x}/90-{x}^{2}/2+\left({e}^{x}x\right)/3-\left(2\mathrm{cos}\left(x\right)\right)/5-\left(6\mathrm{sin}\left(x\right)\right)/5+\left(2{c}_{1}{e}^{x}\right)/3+1/3{c}_{1}{e}^{-2x}+\left({c}_{2}{e}^{x}\right)/3-1/3{c}_{2}{e}^{-2x}-1/2$
Simplify the arbitrary constants: