Answer to Question #157861 in Differential Equations for faseeh

Question #157861

Solve following ODE by method of undetermined coefficients.



(𝑫^πŸβ€•πŸ–π‘«+πŸπŸ“)π’š=πŸ—π’™π’†^πŸπ’™, π’š(𝟎)= πŸ“,π’šβ€²(𝟎)=𝟏𝟎


1
Expert's answer
2021-01-26T04:00:53-0500

Solution

First let’s find fundamental solutions as solution of homogeneous equation.

Y’’ - 8Y’ + 15 = 0

Characteristic equation for it is Ξ»2-8 Ξ» +15=0.Β Β =>Β Β Ξ»1=3, Ξ»2=5

Therefore solution for Y(x) is Y = C1e3x + C2e5xΒ Β where C1 and C2 are arbitrary constants.

Solution of nonhomogeneous equation let’s find as Yp(x)=(Ax+B)e2xΒ Β 

Substitution in equation gives

Ypβ€˜=(A+2Ax+2B)e2x ,Β Ypβ€˜β€™=(2A+2A+4Ax+4B)e2x = 4(A+Ax+B)e2xΒ Β =>Β Β 

[4(A+Ax+B)-8(A+2Ax+2B)+15(Ax+B)] e2x = 9xe2x

(-4A+3B+3Ax) e2x = 9xe2x

3A = 9, 3B-4A = 0Β Β =>Β A = 3, B = 4Β Β =>Β Yp(x)=(3x+4)e2x

So y(x) = Y(x)+Yp(x) = C1e3x + C2e5x +(3x+4)e2xΒ 

From initial conditions C1 + C2 + 4 = 5, 3C1 + 5C2 +11 =10Β Β =>Β Β C1 + C2 = 1, 3C1 + 5C2 = -1Β Β =>Β Β Β 

2C1 = 6, C1 = 3, C2 = 1 - C1 = -2Β =>

y(x) = 3e3x - 2e5x +(3x+4)e2xΒ 

Answer

y(x) = 3e3x - 2e5x +(3x+4)e2x

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