So hard to read ><
Is it ?
If so, it's first order linear, so use the Integrating Factor method.
OK, then read very carefully, for I will post this only once...
To understand the Integrating Factor method, you need to have a basic idea of the Product Rule, i.e. if are functions of , then .
Now examine a general first-order linear ODE:
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What we aim to do is to multiply both sides of the DE by some function of , call it , so that the LHS is a Product-Rule expansion, which can then be written as a derivative, and therefore be integrated. BTW is known as the Integrating Factor (hence the name of the method)
So multiplying both sides by gives
.
The LHS will be a Product Rule expansion if (Can you see why?)
This is a separable ODE, so solving for we have
(You can omit modulus signs and constants because any scalar multiple of the Integrating Factor will do)
.
So, if we multiply both sides of a first-order linear ODE by , we create a Product Rule expansion on the LHS which we can write as a single derivative and integrate both sides. Watch what happens...
which solves the ODE.
Looks complicated in theory, in practice, extremely easy. Look at your problem...
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Here your , so the Integrating Factor is .
Multiplying both sides by the Integrating Factor gives
where is the Integration Constant
.
It doesn't. You must have misunderstood the question. is a solution to that equation only for . Since you have NOT taken a course in differential equations, I suspect you were NOT intended to actually solve that equation. I suspect that the problem was really "Find a value of A such that satisfies the equaton ".
And you can do that just by differentiating the given function and plugging into the equation: if then so the equation becomes . That is the same as and since an exponential is never negative, we can divide both sides of the equation by to get or .