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In this problem, we are going to solve the following initial value problem using the method of laplace transform.
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We have y' ' plus y equal to cosine t, y0 equal to a, and y '0 equal to b.
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Let us start with the laplace transforms.
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We have a capital y as the laplace transform of this unknown function.
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And then for the second derivative, we have s squared y minus sy0 minus y '0.
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So we have s squared y minus as minus b.
00:45
Then for the cosine, which is a well -known function, we have s over s squared plus 1.
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Now putting everything into the laplace transform of this equation, we obtain s squared y minus as minus b plus y equal to s over s squared plus 1.
01:10
And solving this equation for capital y, we obtain b plus a plus b times s plus b s squared plus as cubed divided by 1 plus s squared squared.
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Now we need to perform a partial fraction decomposition on this guy.
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So we have as plus b over 1 over s squared plus cs plus d over 1 over s squared squared.
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Let's set the denominators equal to each other plus cs plus d.
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And this expression should match the numerator of the right -hand side.
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So we have from left b plus d plus a plus c times s plus b s squared plus as cubed.
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So this is equal to b.
02:25
This is a plus 1...