Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients) Find the

Question

Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients) Find the

Kaiya Hardin 2022-04-30 Answered

Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients)
Find the general solution of the following Differential equation

y^{″} - 2 y^{'} + 10 y = e^{x} \cos (3 x)

. We know that the general solution for 2nd order Nonhomogeneous differential equations is the sum of

y_{p} + y_{c}

where

y_{c}

is the general solution of the homogeneous equation and

y_{p}

the solution of the nonhomogeneous. Therefore

y_{c} = e^{x} (c_{1} \cos (3 x) + c_{2} \cos (3 x))

. Now we have to find

y_{p}

. I know in fact that

y_{c} = e^{x} (c_{1} \cos (3 x) + c_{2} \cos (3 x))

. Now we have to find yp. I know in fact that

y_{p} = e^{\times} \frac{\sin (3 x)}{6}

but i do not know how to get there.

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Answers (2)

j3jell5

Answered 2022-05-01 Author has 17 answers

Step 1
since the RHS is a soln of the homogeneous eqn, we can try x multiplied by it. let

y_{p}

be a linear combination of

x e^{x} \cos (3 x)

and

x e^{x} \sin (3 x)

.

y_{p} = a x e^{x} \cos (3 x) + b x e^{x} \sin (3 x)

Step 2
Let

E = D^{2} - 2 D + 10

, where D is the derivative operator.

E [y_{p}] = - 6 a e^{x} \sin (3 x) + 6 b e^{x} \cos (3 x)

E [y_{p}] = e^{x} \cos (3 x)

equating coeffs,

a = 0 and 6 b = 1

y_{p} = \frac{1}{6} x e^{x} \sin (3 x)

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Bethany Mills

Answered 2022-05-02 Author has 14 answers

Step 1
After a lot of trial and error i finnaly understand the strategy behind this type of differential equation. So here's is what i did in order to find the solution (I will provide as much information as i can) In order to calculate a nonhomogeneous differntial equation we must first find the general solution of the homogeneous. So

y^{″} - 2 y^{'} + 10 y = o

(1) using the Auxiliary equation we can easily find that the general solution is

y_{h} = e^{x} (C_{1} \cos (3 x) + C_{2} \sin (3 x)

. Now as cineel mentioned above we can try

y_{h}

multiplied by x. Thus

y_{p} = A x e^{x} \cos (3 x) + B x e^{x} \sin (3 x)

. Now we have to differentiate

y_{p}

two times.
So

y_{p} = A x e^{x} \cos (3 x) + B x e^{x} \sin (3 x)

(1)

y_{p}^{'} = A [e^{x} \cos (3 x) + x e^{x} \cos (3 x) - x e^{x} 3 \sin (3 x)] + B [e^{x} \sin (3 x) + x e^{x} \sin (3 x) + 3 x e^{x} \cos (3 x)

(2)

y_{p}^{″} = A [- 8 x e^{x} \cos (3 x) - 6 e^{\times} \sin (3 x) + 2 e^{x} \cos (3 x) - 3 e^{x} \sin (3 x)] + B [- 8 x e^{x} \sin (3 x) + 6 e^{\times} \cos (3 x) + 2 e^{x} \sin (3 x) + 6 e^{x} \cos (3 x)]

(3)
Step 2
Therefore substituting (1)(2)(3) in our original differential equation we end up with

- 3 A e^{x} \sin (3 x) + 6 B e^{x} \cos (3 x) = e^{x} \cos (3 x)

So

A = 0 and B = \frac{1}{6}

Thus (1) Now becomes

y_{p} = \frac{1}{6} x e^{x} \sin (3 x)

So

y_{p} + y_{h} = e^{x} [C_{1} \cos (3 x) + C_{2} \sin (3 x)] + \frac{1}{6} e^{\times} \sin (3 x)

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