Chapter 6: Q11E (page 332)

Find a general solution for the differential equation with x as the independent variable:
$y^{(4)} + 4 y''' + 6 y'' + 4 y' + y = 0$

Short Answer

Expert verified

The general solution for the differential equation with x as the independent variable is $y = C_{1} e^{- x} + C_{2} x e^{- x} + C_{3} x^{2} e^{- x} + C_{4} x^{3} e^{- x}$

Step by step solution

Auxiliary equation:

The given differential equation is $y^{(4)} + 4 y''' + 6 y'' + 4 y' + y = 0$ . To solve this equation, we look at its auxillary equation which is $m^{4} + 4 m^{3} + 6 m^{2} + 4 m + 1 = 0$ .

By binomial theorem, it is clear seen that the auxillary equation is equal to ${(m + 1)}^{4}$ . So, $m = - 1, - 1, - 1, - 1$ . In other words, -1 is a multiple root repeasted four times.

General solution:

The general solution to the given differential equation is given by

$y = C_{1} e^{- x} + C_{2} x e^{- x} + C_{3} x^{2} e^{- x} + C_{4} x^{3} e^{- x}$ , where $C_{i} (1 \leq i \leq 4)$ are arbitrary constant.

The general solution of the given differential equation is $y = C_{1} e^{- x} + C_{2} x e^{- x} + C_{3} x^{2} e^{- x} + C_{4} x^{3} e^{- x}$

Hence the final solution is $y = C_{1} e^{- x} + C_{2} x e^{- x} + C_{3} x^{2} e^{- x} + C_{4} x^{3} e^{- x}$

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Find a general solution for the differential equation with x as the independent variable:
$y^{(4)} + 4 y''' + 6 y'' + 4 y' + y = 0$

Short Answer

Step by step solution

Auxiliary equation:

General solution:

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Find a general solution for the differential equation with x as the independent variable:y(4)+4y'''+6y''+4y'+y=0

Short Answer

Step by step solution

Auxiliary equation:

General solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

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Study anywhere. Anytime. Across all devices.

Find a general solution for the differential equation with x as the independent variable:
$y^{(4)} + 4 y''' + 6 y'' + 4 y' + y = 0$