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Problem 7: Verify the solution to a N1-ODE
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Given: A N1-ODE and its solution

 * A nonlinear, 1st order ordinary differential equation(N1-ODE):


 * The solution to the N1-ODE:

Find: Verify the solution to the equation
Verify that ($$) is indeed the solution for the N1-ODE ($$).

Solution

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This solution was prepared without referring to previous solutions.
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Basic train of thought is to produce ($$) from ($$).

First we obtain ($$) from ($$):

Then make differential of ($$) with respect of $$x$$, we obtain:

Obviously, ($$) is indeed ($$). Therefore the solution is verified.

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Problem 10: Find the integrating factor and solution of certain non-homogenous L1-ODE-VC
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Given: A non-homogenous L1-ODE-VC
The non-homogenous linear 1st-order ordinary differential equation with varying coefficients(L1-ODE-VC):

Find: Get the integrating factor and solution to the equation
Using Euler Integrating Factor Method(IFM), obtain the integrating factor and solution to ($$).

Solution: Using IFM

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This solution was produced without referring to previous solutions. Basic train of thought: To solve non-homogenous L1-ODE-VC like ($$), the ultimate goal is to reduce the order of the equation. However, first we need to check whether or not the equation satisfies the Two Exactness Conditions, that is:
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 * 1st Condition:


 * 2nd Condition:

If the 1st condition isn't satisfied, then we can transform the equation into the form of ($$) using algebraic method. Then if the 2nd condition isn't satisfied, we could adopt Euler Integrating Factor Method(IFM).

Based on the methodology above, the equation will be solved below: First put the right-hand side of ($$) to the left-hand side, we obtain:

Thus ($$) is transformed into form of ($$).

Then introduce $$h(x)$$ to ($$) such that:(Here we neglect $$h(y)$$ situation which will leads to contradictory result.)

Apply 2nd exactness condition ($$):

We obtain:

Thus:

Make integration on both sides of the equation ($$), we obtain:

thus:

where $$k, K$$ are integration constants and $$K\neq 0$$.

Institute ($$) into ($$), we obtain:

Reduce $$K$$, we get:

Again, make integration on($$), thus:

where $$C$$ is an integration constant.

To put in a nutshell, the solution to the problem is ($$)&($$).

Discussion: how come there are TWO constants in the process but finally there is only ONE?
It should be noted that $$h(x)$$ we just produced is actually a function group $$h(x)=K\cdot x$$ which varies with $$K(K\neq 0)$$. However, $$k$$ will be reduced when instituted into the equation($$). Therefore, $$K$$ will not affect the result of $$y(x)$$ and the fact that there is only ONE integration constant $$C$$ in the result, which is corresponding to FIRST ORDER.