A first-order equation Consider the equation t² y′(t) + 2ty(t)...

Question

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗ

c. Find the solution that satisfies the condition y(1) = 0

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the solution to the equation Squared multiplied by W S, plus 2s multiplied by W of S is equal to E raised to the quantity of minus 2s in quantity with W2 equal to 1 after writing the left hand side as a derivative of a single term. So the first step is to rewrite the left hand side of our equation as a derivative of a single term. If we look at the left hand side of our equation, we notice that this appears to be the result of a product rule. And so, if you look at the derivative with respect to S, Of the quantity of Squared multiplied by W of S. In quantity that this is going to be equal to. Squared multiplied by W of S. Plus 2 S multiplied by W of S, which is exactly what we have on the left hand side of our equation. That means that we can rewrite our equation as the derivative with respect to S of the quantity of Squad, multiplied by W of S. In quantity, and this is going to be equal to E raised to the quantity of minus 2 S in quantity. So the next step is to integrate both sides of the equation with respect to S. And so when we integrate the left hand side, we just get S squared multiplied by W of S. And this is going to be equal to, and when we integrate the right hand side of our equation, we get -1 divided by 2, multiplied by E, raised to the quantity of minus 2 S in quantity. But each of these integrations will have an integration constant, so we can combine the two integration constants, one from each side of the equation, into a single integration constant of plus C on the right hand side of our equation. And so now we need to solve for WFS. So to do that, we'll divide both sides of the equation by Squad, and this will give us W of S. is equal to the quantity of C minus 1/2 multiplied by E raised the quantity of minus to S in quantity, and that whole quantity is going to be divided by Squared. Now we need to use our initial condition in order to find our value for C. and we're told that WF 2 is equal to 1. So whenever S is equal to 2, WFS is going to be equal to 1. So we can make those substitutions into our expression here, and we'll see that 1 is going to be equal to the quantity of C minus 1/2 multiplied by E raised to the quantity of minus 2 multiplied by 2. In quantity, and that whole quantity is going to be divided by. 2 squared. So now we need to solve for C. So we'll multiply both sides of the equation by 4, since 22 is 4. In doing so, we'll get that 4 is equal to C minus 1/2 multiplied by E raised to the minus 4. And so now we just need to add 1/2 multiplied by E raised to the minus 4 power to both sides of the equation. In doing so, we'll get that C is equal to 4 plus 1/2 multiplied by E raised to the minus 4 power. So now we can make this substitution back into our expression for W of S. In doing so, we'll get. WFS is equal to The quantity of 4. Plus, 1/2 multiplied by E raised to the minus 4 power. Minus One half multiplied by E raised to the quantity of minus 2 S in quantity, and that whole quantity is going to be divided by Squared. And we can multiply both the numerator and denominator by 2 to remove the fraction out of the numerator. In doing so, we'll see that W of S. is going to be equal to the quantity of 8. Plus, he raised to the -4 power. Minus E rates to the quantity of minus 2 S in quantity, and that whole quantity is going to be divided by the quantity of 2 S squad in quantity. And this here is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗc. Find the solution that satisfies the condition y(1) = 0

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A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗ
c. Find the solution that satisfies the condition y(1) = 0