Math / Calculus

Question1. Find the general solution of the following differential equation by the method of undetermined coefficients: $y^{\prime \prime}+2 y^{\prime}-24 y=16-(x+2) e^{4 x}$

Studdy Solution

STEP 1

What is this asking? We need to find all the solutions $y(x)$ that satisfy a given differential equation, which involves second derivatives, first derivatives, and the function itself.
The right-hand side has a constant and an exponential term multiplied by $x$ , so we'll need to be clever! Watch out! Don't forget to consider both the complementary solution and the particular solution.
Also, make sure to adjust the form of your particular solution if any part of it overlaps with the complementary solution.

STEP 2

1. Find the complementary solution.
2. Find the particular solution.
3. Combine solutions for the general solution.

STEP 3

First, let's find the complementary solution $y_c$ , which is the solution to the homogeneous equation: $y^{\prime \prime} + 2y^{\prime} - 24y = 0.$ To do this, we'll guess a solution of the form $y_c = e^{rx}$ .

STEP 4

Plugging our guess into the homogeneous equation, we get: $r^2 e^{rx} + 2re^{rx} - 24e^{rx} = 0.$ Since $e^{rx}$ is never zero, we can divide both sides by $e^{rx}$ to get the characteristic equation: $r^2 + 2r - 24 = 0.$

STEP 5

Now, we solve for $r$ .
This quadratic equation factors nicely as $(r+6)(r-4) = 0$ , so the roots are $r_1 = -6$ and $r_2 = 4$ .

STEP 6

Therefore, the complementary solution is: $y_c = C_1 e^{-6x} + C_2 e^{4x},$ where $C_1$ and $C_2$ are arbitrary constants.

STEP 7

Now, let's find the particular solution $y_p$ .
Since the right-hand side of our original equation is $16 - (x+2)e^{4x}$ , we'll guess a particular solution of the form: $y_p = A + (Bx + C)e^{4x}.$

STEP 8

Uh oh!
Notice that the $e^{4x}$ term in our guess overlaps with the $e^{4x}$ term in our complementary solution.
To fix this, we'll multiply the overlapping part by $x$ , giving us a new guess: $y_p = A + (Bx^2 + Cx)e^{4x}.$

STEP 9

Now, we calculate the first and second derivatives of $y_p$ : $y_p^{\prime} = (2Bx + C)e^{4x} + 4(Bx^2 + Cx)e^{4x} = (4Bx^2 + (2B+4C)x + C)e^{4x}$ $y_p^{\prime \prime} = (8Bx + (2B+4C))e^{4x} + 4(4Bx^2 + (2B+4C)x + C)e^{4x} = (16Bx^2 + (16B + 8C)x + 2B + 8C)e^{4x}.$

STEP 10

Substitute $y_p$ , $y_p^{\prime}$ , and $y_p^{\prime \prime}$ into the original differential equation: $(16Bx^2 + (16B+8C)x + 2B+8C)e^{4x} + 2(4Bx^2 + (2B+4C)x + C)e^{4x} - 24(A + (Bx^2 + Cx)e^{4x}) = 16 - (x+2)e^{4x}.$

STEP 11

Simplify and group like terms: $-24A + (16B + 8B - 24B)x^2 e^{4x} + (16B + 8C + 4B + 8C - 24C)xe^{4x} + (2B + 8C + 2C)e^{4x} = 16 - xe^{4x} - 2e^{4x}.$ $-24A + (20B)xe^{4x} + (2B + 10C)e^{4x} = 16 - xe^{4x} - 2e^{4x}.$

STEP 12

Equate coefficients: $-24A = 16 \implies A = -\frac{2}{3}$ $20B = -1 \implies B = -\frac{1}{20}$ $2B + 10C = -2 \implies -\frac{1}{10} + 10C = -2 \implies 10C = -\frac{19}{10} \implies C = -\frac{19}{100}.$

STEP 13

So, the particular solution is: $y_p = -\frac{2}{3} - \left(\frac{1}{20}x^2 + \frac{19}{100}x\right)e^{4x}.$

STEP 14

Finally, the general solution is the sum of the complementary and particular solutions: $y = y_c + y_p = C_1 e^{-6x} + C_2 e^{4x} - \frac{2}{3} - \left(\frac{1}{20}x^2 + \frac{19}{100}x\right)e^{4x}.$

STEP 15

The general solution to the given differential equation is $y(x) = C_1 e^{-6x} + C_2 e^{4x} - \frac{2}{3} - \left(\frac{1}{20}x^2 + \frac{19}{100}x\right)e^{4x}.$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms