Question

Using the method of undetermined coefficients find the general solution of the differential equation
y^iv-2y"'+2y"=3e^-x+3e^-x+e^-xsinx

Answer 1

Using the method of undermined coefficients, find the general solution of the differential equation y^iv-2y'''+2y''=3e^-x+2e^-xx+e^-x sin x

$y iv −2y ′′′ +2y ′′ =3e ^ {−x} +2e ^ {−x} x+e ^ {−x} sinx$

To determine the general solution needed find a complementary and partial solutions. First, determine a partial solution.

Rewrite the right part of the equation as

$f(x)=(3+2x)e^ {−x} +e^ {−x} sinx$

Therefore, a partial solution of the equation have a form

$y_ p ​ (x)=(A+Bx)e ^ {−x} +e ^ {−x} (Csinx+Dcosx)$

Find the derivatives.

$y_ p ′ ​ =Be^ {−x} −(A+Bx)e^ {−x} −e ^ {−x} ((C+D)sinx−(C−D)cosx)$

$y_ p ′′ ​ =−2Be^ {−x} +(A+Bx)e^ {−x} +e^ {−x} (2Dsinx−2Ccosx)$

$y_ p ′′′ ​ =3Be^ {−x} −(A+Bx)e^ {−x} +e^ {−x} ((2C−2D)sinx+(2C+2D)cosx)$

$y_ p iv ​ =−4Be^ {−x} +(A+Bx)e^ {−x} −e^ {−x} (4Csinx+4Dcosx)$

Substituting these into the equation, obtain

$−14Be ^ {−x} +5(A+Bx)e^ {−x} +e^ {−x} ((8D−8C)sinx−(8D+8C)cosx) =3e^ {−x} +2xe^ {−x} +e^ {x} sinx$

Rewrite the left part.

$(5A−14B)e ^ {−x} +5Bxe ^ {−x} +e^ {−x} ((8D−8C)sinx−(8D+8C)cosx) =3e^ {−x} +2xe^ {−x} +e^ {x} sinx$

Therefore,

$5A−14B=3$

$5B=2$

$8D−8C=1$

$−8D−8C=0$

Solving these linear system, get that $A= {43 \above{2pt} 25} ​ ,B= 2/5 ​ ,C=−1/ 16 ​$ and $D=1/16$

Thus, the partial solution is

$y_ p ​ (x)=( {43 \above{2pt} 25} ​ + {2 \above{2pt} 5} ​ x)e^ {−x} +e^ {−x} ({1 \above{2pt} 16} ​ cosx− {1 \above{2pt} 16} ​ sinx)$

Now, determine a complementary solution. Solve the characteristic equation.

$λ^ 4 −2λ^ 3 +2λ ^ 2 =0$

$λ^ 2 (λ^ 2 −2λ+2)=0$

This equation have the roots $λ=0$ (multiplicity 2) and $λ=1±i$ .Therefore, the complementary solution is

$y c ​ (x)=C_ 1 ​ +C _ 2 ​ x+C _ 3 ​ e ^ {x} cosx+C_ 4 ​ e ^ {x} sinx$

So, the general solution is

$y(x)=$ $C_ 1 ​ +C _ 2 ​ x+C _ 3 ​ e ^ {x} cosx+C_ 4 ​ e ^ {x} sinx$ +$( {43 \above{2pt} 25} ​ + {2 \above{2pt} 5} ​ x)e^ {−x} +e^ {−x} ({1 \above{2pt} 16} ​ cosx− {1 \above{2pt} 16} ​ sinx)$