Answer to Question #122597 in Differential Equations for jse

Question

1. Solve the given differential equation by undetermined coefficients.
y'' + 3y = −48x^2 (e^3x).

y(x) =____


2. Find a particular solution of the given differential equation. Use a CAS as an aid in carrying out differentiations, simplifications, and algebra.

y^(4) + 2y'' + y = 1 cos x − 3x sin x

yp = _____

Answer 1

1.y'' + 3y = −48x²e^3x.

Solution:

y'' + 3y=0

$\lambda$²+3 =0

$\lambda$ ₁=-i3^1/2, $\lambda$ ₂=i3^1/2.

y₀(x)=c₁cos(3^1/2 x)+c₂sin(3^1/2x),

y_p(x)=(Ax²+Bx+C)e^3x

y_p'(x)=(3Ax²+(3B+2A)x+3C+B)e^3x

y_p''(x)=(9Ax²+(9B+12A)x+2A+6B+9C)e^3x

x²e^3x:3A+9A=-48,

xe^3x:3B+9B+12A=0,

e^3x:3C+2A+6B+9C=0.

=>A=-4,B=4,C=-4/3.

y_p(x)=(-4x²+4x-4/3)e^3x,

y(x)=y₀(x)+y_p(x)

Answer:

y(x)=c₁cos(3^1/2 x)+c₂sin(3^1/2x)+(-4x²+4x-4/3)e^3x

2.y⁽⁴⁾ + 2y'' + y = cos x − 3x sin x 

Solution:

y⁽⁴⁾ + 2y'' + y =0

$\lambda$ ⁴+2$\lambda$ ²+1=0

($\lambda$ ²+1)²=0

$\lambda$ ₁=-i, k=2,

$\lambda$₂ =i, k=2

y_p(x)=(Ax³+Bx²)cos(x)+(Cx³+Dx²)sin(x)

y_p'(x)=...,y_p''(x)=...y_p'''(x)=...,y⁽⁴⁾(x)=...

y_p⁽⁴⁾ + 2y_p'' + y_p=cos(x)(-24Ax-8B+24C)+sin(x)(-24A-24Cx-8D)

-24A=0

-8B+24C=1

-24C=3

-24A-8D=0

=>A=0,B=1/4, C=1/8, D=0

Answer:

y_p(x)=x²/4cos(x)+x³/8sin(x)