Solve the following initial value problem: with So I isolated the y and it became 8y=(9-dy/dt)t how would I finish the rest?
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Originally Posted by superman69 Solve the following initial value problem: with So I isolated the y and it became 8y=(9-dy/dt)t how would I finish the rest? Divide both sides by t: . Now use the integrating factor technique.
You can also do this as a "homogeneous" equation. Write it as . Let . Then so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation.
Originally Posted by mr fantastic Divide both sides by t: . Now use the integrating factor technique. What is the integrating factor technique? I did not learn that yet.
Originally Posted by HallsofIvy You can also do this as a "homogeneous" equation. Write it as . Let . Then so y'= tu'+ u and the equation becomes tu'+ u= 9- 8u and so tu'= 9- 9u, a separable equation. So now I use the equation tu'= 9- 9u and plug in u' and u?
Originally Posted by superman69 What is the integrating factor technique? I did not learn that yet. . The Integrating Factor is . So multiply both sides by the integrating factor to get Can you see that the LHS is a product rule expansion of ? . Now use the initial condition to find .
Originally Posted by superman69 So now I use the equation tu'= 9- 9u and plug in u' and u? The equation is . It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's).
Originally Posted by mr fantastic The equation is . It's seperable and your task is to solve for u and so get y from y = ut (and I hope you have learnt about seperable DE's). It helps if you write the DE as so that when you take the integral of both sides with respect to t . I'm sure you can go from here.
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