7–16. Verifying general solutions Verify that the given functi...

Question

7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.

u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³

Accepted Answer

Welcome back, everyone. Verify that the function V X equals A X 6 plus B X 3 minus x 4th is a solution to the differential equation X 2 V X minus 30 X equals 18 x 4th, where A and B are arbitrary constants. A yes, V of X is a solution, and B, no V of X is not a solution. Our differential equation contains V of X and V of X. So let's identify the second derivative by. Identifying the first derivative to begin with. The first derivative of B of X is going to be the derivative of A X to the 6th, plus B X 3 minus X 4th. We can apply the power rule, right, and we get A multiplied by 6 X to the 5th, plus. Now, the derivative of X to the 3 is going to be -3 X to the 4th, so we get -3BX to the negative4th, and the derivative of X the 4th is going to be 3 X. I'm sorry, 4 X cubed, right? So that'd be 4 X cubed. So we have our first derivative. We can write it as 6 A X to the 5th minus 3 B X 4 minus 4 X cubed. And now, because we want to identify the second derivative, we're going to differentiate this result. Using the power rule, we get 5 multiplied by 6, that's 30, A X to the 4th, right? minus. 3 multiplied by -4 is -12, so we are subtracting -12 for basically adding 12B X to the negative 5th. Minus 4 multiplied by 3 is 12. And then we have x squared, so our second derivative. Is 30AX to the 4th plus 12BX to the negative 5th minus 12 X squad. Now what we're going to do is simply substitute this result into our differential equation. We get X squared multiplied by 30, A X to the 4th plus 12 B X to the negative 5th. Minus 12 x squad. And then we're subtracting 30 B of X. So 30 is multiplied by A X to the 6th. Plus B X 3 minus X 4th. We're going to simplify this expression and check if we get 18 X to the 4th. If we do get 18 X to the 4th, then we can conclude that V of X is indeed a solution. Otherwise, it is not. So now let's go ahead and distribute. We're going to get 30AX to the 6th, plus 12BX to the negative 3, minus 12 X to the 4th, minus 30A X to the 6th, minus 30BX to the negative 3, plus 30 X to the 4th. Let's see if we can cancel out any of the terms, we can cancel out 30 AX to the 6th, because we have a positive term and a negative term. Then we have 12 BX to the negative 3, minus 30 BX to the negative 3, which gives us -18. BX to the negative 3. And then we are going to take -12 X to the 4th and add 30 X to the 4th, which gives us 18 X to the 4th. So we indeed have that term, 18 X to the 4th, but we have an extra term, which is -18 BX to the negative 3, right? So this is not equal to 18 X to the 4th. And therefore we can conclude that the correct answer is B. No, VX is not a solution. Thank you for watching.

7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³

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7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
u(t) = C₁t⁵ + C₂t⁻⁴ - t³; t²u''(t) - 20u(t) = 14t³