Answer to Question #112246 in Differential Equations for Utkarsh mishra

Question #112246
Using the method of undetermined coefficients find the general solution of the differential equation
y^iv-2y"'+2y"=3e^-x+3e^-x+e^-xsinx
1
Expert's answer
2020-04-28T17:31:55-0400

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 \niv\n \u22122y \n\u2032\u2032\u2032\n +2y \n\u2032\u2032\n =3e ^\n{\u2212x}\n +2e ^\n{\u2212x}\n x+e ^\n{\u2212x}\n 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^ \n{\u2212x}\n +e^ \n{\u2212x}\n sinx"


Therefore, a partial solution of the equation have a form

"y_ \np\n\u200b\t\n (x)=(A+Bx)e ^\n{\u2212x}\n +e ^\n{\u2212x}\n (Csinx+Dcosx)"

Find the derivatives.

"y_ \np\n\u2032\n\u200b\t\n =Be^ \n{\u2212x}\n \u2212(A+Bx)e^ \n{\u2212x}\n \u2212e ^\n{\u2212x}\n ((C+D)sinx\u2212(C\u2212D)cosx)"

"y_ \np\n\u2032\u2032\n\u200b\t\n =\u22122Be^ \n{\u2212x}\n +(A+Bx)e^ \n{\u2212x}\n +e^ \n{\u2212x}\n (2Dsinx\u22122Ccosx)"

"y_ \np\n\u2032\u2032\u2032\n\u200b\t\n =3Be^ \n{\u2212x}\n \u2212(A+Bx)e^ \n{\u2212x}\n +e^ \n{\u2212x}\n ((2C\u22122D)sinx+(2C+2D)cosx)"

"y_ \np\niv\n\u200b\t\n =\u22124Be^ \n{\u2212x}\n +(A+Bx)e^ \n{\u2212x}\n \u2212e^ \n{\u2212x}\n (4Csinx+4Dcosx)"

Substituting these into the equation, obtain

"\u221214Be ^\n{\u2212x}\n +5(A+Bx)e^ \n{\u2212x}\n +e^ \n{\u2212x}\n ((8D\u22128C)sinx\u2212(8D+8C)cosx)\n=3e^ \n{\u2212x}\n +2xe^ \n{\u2212x}\n +e^ \n{x}\n sinx"

Rewrite the left part.

"(5A\u221214B)e ^\n{\u2212x}\n +5Bxe ^\n{\u2212x}\n +e^ \n{\u2212x}\n ((8D\u22128C)sinx\u2212(8D+8C)cosx)\n=3e^ \n{\u2212x}\n +2xe^ \n{\u2212x}\n +e^ \n{x}\n sinx"

Therefore,

"5A\u221214B=3"

"5B=2"

"8D\u22128C=1"

"\u22128D\u22128C=0"

Solving these linear system, get that "A= {43 \\above{2pt} 25}\n\u200b\t\n ,B= 2\/5\n\n\u200b\t\n ,C=\u22121\/ \n16\n\n\u200b" and "D=1\/16"

Thus, the partial solution is

"y_ \np\n\u200b\t\n (x)=( {43 \\above{2pt} 25}\n\n\u200b\t\n + {2 \\above{2pt} 5}\n\n\u200b\t\n x)e^ \n{\u2212x}\n +e^ \n{\u2212x}\n ({1 \\above{2pt} 16}\n\n\u200b\t\n cosx\u2212 \n{1 \\above{2pt} 16}\n\u200b\t\n sinx)"

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

"\u03bb^ \n4\n \u22122\u03bb^ \n3\n +2\u03bb ^\n2\n =0"

"\u03bb^ \n2\n (\u03bb^ \n2\n \u22122\u03bb+2)=0"


This equation have the roots "\u03bb=0" (multiplicity 2) and "\u03bb=1\u00b1i" .Therefore, the complementary solution is

"y \nc\n\u200b\t\n (x)=C_ \n1\n\u200b\t\n +C _\n2\n\u200b\t\n x+C _\n3\n\u200b\t\n e ^\n{x}\n cosx+C_ \n4\n\u200b\t\n e ^\n{x}\n sinx"


So, the general solution is

"y(x)=" "C_ \n1\n\u200b\t\n +C _\n2\n\u200b\t\n x+C _\n3\n\u200b\t\n e ^\n{x}\n cosx+C_ \n4\n\u200b\t\n e ^\n{x}\n sinx" +"( {43 \\above{2pt} 25}\n\n\u200b\t\n + {2 \\above{2pt} 5}\n\n\u200b\t\n x)e^ \n{\u2212x}\n +e^ \n{\u2212x}\n ({1 \\above{2pt} 16}\n\n\u200b\t\n cosx\u2212 \n{1 \\above{2pt} 16}\n\u200b\t\n sinx)"


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