Differential Equation using Integrating Factor

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The discussion revolves around solving the ordinary differential equation (ODE) using the integrating factor method. The initial equation is transformed to facilitate the application of this method, leading to the identification of the integrating factor as (1+x). After applying the integrating factor and rearranging, the solution is derived, resulting in y = c/(1+x). The participants confirm the correctness of the steps taken in the solution process. The final solution effectively demonstrates the application of the integrating factor method to the given ODE.
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Homework Statement



The ODE is

(1+x)*dy/dx - xy = x+x^2

Homework Equations



The method of solution is to be through the use of the integration factor.

The Attempt at a Solution



First, I divided each side by (1+x) to produce

dy/dx - xy/(1+x) = x

then factor out the x on the LHS to produce

dy/dx - x * (y/(1+x)) = x

then divide both sides by x to produce

dy/dx - y/(1+x) = 0

now move the -y/(1+x) to the RHS to produce

dy/dx = y/(1+x)

That is all the farther I have progressed. Am I correct so far?

Thanks
Matt
 
Last edited:
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Yes, you are correct so far by what progress is that? How does that help you find the integrating factor?

The left side of your equation is (1+x)(dy/dx)- xy. The whole point of an "integrating factor" is that it is a function u(x) such that u(x)(1+x)(dy/dx)- uxy= d(u(x)(1+x)y)/dx.
Go ahead and use the product rule on the right side. That gives u(1+x)(dy/dx)+ u(1)y+ (du/dx)(1+x)y and that must be equal to u(x)(1+x)(dy/dx)- uxy. Well, the "u(1+x)(dy/dx)" cancels immediately leaving u(x)y+ (1+x)y(du/dx)= -uxy so (1+x)y(du/dx)= -xy(1+ u). Now the "y"s cancel so (1+x)(du/dx)= -x(1+u) which is a separable equation. du/(1+u)= -dx/(x(1+x)). Integrate that to find the integrating factor.
 
By progress, I was meaning the reformatting of the original equation. I believed that this equation is seperable but the assignment was to use the integrating factor method.

I would rather take the path of the seperable equation.

Starting with

dy/dx = y/(1+x)

multiplying both sides by dx produces

dy = y/(1+x)dx

dividing both sides by y produces

dy/y = dx/(1+x)

integrating both sides produces

ln(y) = ln(1+x) + c

Is this correct so far?


Now proceeding forward with your instructions to integrate

du/(1+u) = -dx/(x(1+x)

upon integration of both sides

ln(1+u) = -ln(x)+ln(1+x)

now the integrating factor would be

e^(ln(1+u))

upon integration would be

1+u

Am I correct so far?

Thanks
Matt
 
Last edited:
OK. I think I have the solution.

With the integrating factor of 1+x and using the equation

dy/dx -y/(1+x) = 0

multiplying both sides by the integrating factor produces

(1+x)*dy/dx - (1+x)*y/(1+x) = 0

which is equal to

(1+x) * dy/dx - y = 0

upon integrating

(1+x)*y = c

solving for y produces

y = c/(1+x)

Am I correct?

Thanks
Matt
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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