Which is a standard way of getting the left-hand side into the form of the derivative of a product.
Balloon Calculus: standard integrals, derivatives and methods
I'm getting prepared to take my differential equations final on Tuesday and my professor grades extremely harshly and in order to do well on the final and pass the class, I want to make sure I understand exactly what he's doing in his solutions.
Here's the problem, and solution.
Given the first order linear equation x^2(dy/dx) - 3xy = x^6 find the general solution. Write explicitly in terms of y.
Solution from professor:
(dy/dx) - (3/x)y = x^4 -> u(x) = exp(int(-3/x dx)) = exp(-3lnx) = 1/x^3
1/x^3(dy/dx) - (3/x)y = x
d/dx((1/x^3)y) = x
(1/x^3)*y = (x^2)/2 + c
y = (1/2)x^5 + cx^3
The thing I am most confused about is how he knows that u(x) = exp(int(-3/x dx)). I don't have any idea how I am supposed to realize something like that on the final. I know it's u-substitution, but for the life of me I can't figure out how he got there.
Thanks,
Sean
Which is a standard way of getting the left-hand side into the form of the derivative of a product.
Balloon Calculus: standard integrals, derivatives and methods