I've been working on this hw problem for a while now, but I'm stuck in the integration process. I'm pretty sure I made an error, cause I can't seem to be able to integrate the right side of the equation.
Q: (1/(x^(2)+1))y' + xy = 3
using the equation d/dx(ry)=f(x)r(x) I found r(x) to be e^((1/4)x^(4) + (1/2)x^(2)). Plugging it back into the equation I have d/dx(ry)=(3x^(2)+3)(e^((1/4)x^(4) + (1/2)x^(2))). I tried u-substitution to integrate, but it's not working out. Please explain how to go about this problem, I'd like to know where my mistake is. Thanks!
Hmmm. It appears your integrating factor is correct. Evaluating the right-hand side is intractable.
a couple of online solvers both come up with y in terms of that integral, which is not evaluated.
Beats me.
I see, thanks!
To integrate the right side of the equation, we can use integration by parts method. However, before attempting integration by parts, let's simplify the expression a bit.
Starting with your given expression for r(x):
r(x) = e^((1/4)x^4 + (1/2)x^2)
Let's rewrite it as:
r(x) = e^((1/4)x^4) * e^((1/2)x^2)
Now, let's differentiate r(x) and find d/dx(r(x)):
d/dx(r(x)) = d/dx(e^((1/4)x^4) * e^((1/2)x^2))
To differentiate this product, we can apply the product rule:
d/dx(r(x)) = e^((1/4)x^4) * d/dx(e^((1/2)x^2)) + e^((1/2)x^2) * d/dx(e^((1/4)x^4))
Using the chain rule, we can calculate the derivatives of the exponential terms:
d/dx(e^((1/2)x^2)) = (1/2)e^((1/2)x^2) * d/dx(x^2) = (1/2)e^((1/2)x^2) * 2x = xe^((1/2)x^2)
d/dx(e^((1/4)x^4)) = (1/4)e^((1/4)x^4) * d/dx(x^4) = (1/4)e^((1/4)x^4) * 4x^3 = xe^((1/4)x^4)
Substituting these derivatives back into the expression for d/dx(r(x)), we have:
d/dx(r(x)) = xe^((1/4)x^4) * xe^((1/2)x^2) + e^((1/2)x^2) * xe^((1/4)x^4)
Simplifying, we get:
d/dx(r(x)) = (x^2)e^((1/4)x^4 + (1/2)x^2) + (x)e^((1/2)x^2)e^((1/4)x^4)
Now that we have the correct expression for d/dx(r(x)), we can proceed to integrate it. Let me know if you would like further assistance with the integration or any other aspect of the problem.