Help with Initial Value Problems

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SUMMARY

This discussion focuses on solving initial value problems in calculus, specifically two differential equations. The first problem, y' + 5y = t with initial condition y(0) = 0, is solved using a particular solution approach and the homogeneous solution, yielding y = t/5 - 1/25 + Ce^(-5t). The second problem, y' = y/sqrt(t) with initial condition y(1) = -1, is approached through separation of variables, leading to the integration of both sides. The integrating factor technique is also mentioned as an alternative for the first problem.

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vpathak
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Hi, I have a Calculus test coming up and am very confused on how to start approching these types of problems (initial value problems) If possible can someone work these 2 different types of problems out and show some work so I know how to proceed with the rest of them? I don't think they are that hard, for some reason I'm just confused. Thanks


Problem 1: y' + 5y = t, y(0) = 0

Problem 2: y' = y/sqrt(t), y(1) = -1

Thanks again for all those who help!
 
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Problem 1: We first find a particular solution to y' + 5y = t. Suppose y_p = at + b, then the equation becomes

a + (5at + 5b) = t

<=>

{ 5a = 1
{ a + 5b = 0

<=>

{ a = 1/5
{ b = -1/25

Thus y_p = t/5 - 1/25 is a particular solution.

We must then solve the homogenous equation:

y' + 5y = 0.

This is a rather well-studied equation, y_h = Ce^(-5t) is the general solution (it could be solved with an integrating factor also). Thus, the general solution to y' + 5y = t can be written:

y = y_p + y_h = t/5 - 1/25 + Ce^(-5t). You can then use your initial condition to calculate C.

Problem 2: y' = y/sqrt(t) <=> dy/dt = y/sqrt(t) <=> 1/y * dy = 1/sqrt(t) * dt. What happens if you integrate both sides?
 
Just adding to Muzza here:
Problem 1)
If you know about the "integrating factor"-technique, you may use this technique instead to derive the solution.
 

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