Problem - You want to integrate | |
![]() | (1) |
However, that f(z)y term really messes things up! If you only had an expression of the form | |
![]() | (2) |
things would be much easier, then you could integrate with respect to z and find y(z). How can you combine f(z) and y to get this simplification? First note that ![]() | |
![]() | (3) |
That's looking close to the left hand side of equation (1), but there is a "u" in front of the dy/dz term, and a du/dz expression where f(z) is. If you had a form of u such that du/dz = u f(z), then you could manipulate equation (3): | |
![]() | (4) |
where the term in brackets is the left hand side of equation (1). You need du/dz = u f(z). Recall | |
![]() | (5) |
If you define ![]() ![]() | |
![]() | (6) |
This satisfies the condition that du/dz=u f(z). That's the ezpression you needed! Therefore, ![]() | |
![]() | (7) |
where the term in brackets is the left hand side of equation (1). CONCLUSION: If your problem is of the form | |
![]() | (1) |
you can multiply both sides of the equation by the | |
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(which you should be able to evaluate, since you know f(z)), to yield | |
![]() | (9) |
or, substituting from equation (7) | |
![]() | (10) |
so that | |
![]() | (11) |
![]() | (12) |
EXAMPLE | |
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f(t)=k2, so | |
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From equation (12), the solution is then | |
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The constant can be obtained from the intitial condition that at t=0, CB=0; | |
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