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:
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...
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 .