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Find the general solution of y(4) + 2y" + 6y" + 324 + 40y = 0

User Elya
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1 Answer

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To find the general solution of the given differential equation:

y(4) + 2y" + 6y' + 324 + 40y = 0

We can rearrange the equation and combine like terms:

y(4) + 2y" + 6y' + 40y + 324 = 0

Simplifying further, we have:

2y" + 6y' + 44y + 324 = 0

Now, let's solve the homogeneous version of this equation, which is obtained by setting the equation equal to zero:

2y" + 6y' + 44y = 0

To solve this homogeneous linear ordinary differential equation, we assume a solution of the form y = e^(rt), where r is a constant. Substituting this into the equation, we get:

2r^2e^(rt) + 6re^(rt) + 44e^(rt) = 0

Factoring out e^(rt), we have:

e^(rt)(2r^2 + 6r + 44) = 0

For this equation to hold, either e^(rt) = 0 (which is not possible) or 2r^2 + 6r + 44 = 0. Solving the quadratic equation, we find the roots:

r = (-6 ± √(6^2 - 4 * 2 * 44)) / (2 * 2)

r = (-6 ± √(36 - 352)) / 4

r = (-6 ± √(-316)) / 4

Since the discriminant is negative, the roots are complex. Let's write the roots as:

r = (-6 ± √316i) / 4

r = (-3 ± √79i) / 2

The general solution for the homogeneous equation is:

y_h = C1e^(-3t/2)cos(√79t/2) + C2e^(-3t/2)sin(√79t/2)

Now, to find the general solution for the original non-homogeneous equation, we can use the method of undetermined coefficients. We assume a particular solution of the form:

y_p = At + B

Substituting this into the original equation, we have:

2(0) + 6A + 44(At + B) + 324 = 0

Simplifying, we get:

6A + 44At + 44B + 324 = 0

To satisfy this equation, we equate the coefficients of like terms:

44A = 0 => A = 0

6A + 44B + 324 = 0 => 44B = -6A - 324 => B = -3/11

Therefore, the particular solution is:

y_p = (-3/11)t

Finally, the general solution of the non-homogeneous equation is the sum of the homogeneous and particular solutions:

y = y_h + y_p

y = C1e^(-3t/2)cos(√79t/2) + C2e^(-3t/2)sin(√79t/2) - (3/11)t

where C1 and C2 are arbitrary constants.

User Jeremy Baker
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