Find the general solution for the following differential equation. d3ydx3−d2ydx2−dydx+y=x+e−x.

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ankarskogC

Answered question

2021-11-05

Find the general solution for the following differential equation.
$\frac{d^{3} y}{{d x}^{3}} - \frac{d^{2} y}{{d x}^{2}} - \frac{d y}{d x} + y = x + e^{- x}$ .

Answer & Explanation

Alara Mccarthy

Skilled2021-11-06Added 85 answers

Step 1
The ordinary differential equation involves the derivative of a unknown with respect to one dependent variable and one independent variable. ordinary differential equation does not include partial derivatives.
Step 2
The auxiliary equation is,

D^{3} - D^{2} - D + 1 = 0

D=-1,1,1
So, the complimentary function is,

C . F = A e^{- x} + B e^{x} + C x e^{x}

Particular integral is,

P . I = \frac{1}{D^{3} - D^{2} - D + 1} x + e^{- x}

= \frac{1}{D^{3} - D^{2} - D + 1} x + \frac{1}{D^{3} - D^{2} - D + 1} e^{- x}

= {[1 + (D^{3} - D^{2} - D)]}^{- 1} x + x \frac{1}{3 D^{2} - 2 D - 1} e^{- x}

= [1 - (D^{3} - D^{2} - D) + {(D^{3} - D^{2} - D)}^{2} \dots . .] x + x \frac{1}{3 {(- 1)}^{2} - 2 (- 1) - 1} e^{- x}

= x - 1 + \frac{x e^{- x}}{4}

The general solution is,

y (x) = A e^{- x} + B e^{x} + C x e^{x} + x - 1 + \frac{x e^{- x}}{4}

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