Originally Posted by
TwistedOne151 Sorry for the notation, but LaTeX appears to be mostly down.
The endpoints of L are (0,1) and (t,y(t)). Now, we can use the fact that the curve L is given by y as a function y(x) of x to take:
dy=(dy/dx)dx=y'(x)dx, and so our equation then becomes:
∫_{1}^{t}(-5x^3+3y)dx+xy'dx=(t^4-1)/4
∫_{1}^{t}(-5x^3+3y+xy')dx=(t^4-1)/4.
(∫_{1}^{t} means definite integral from 1 to t).
Now, we differentiate both sides with respect to t. Using the fundamental theory of calculus on the left hand side, we get:
(d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)[(t^4-1)/4].
-5t^3+3y(t)+ty'(t)=t^3
ty'(t)+3y(t)=6t^3.
Or, renaming the independent variable back to x,
xy'(x)+3y(x)=6x^3.
This is a nonhomogenous first-order linear differential equation, and should probably be solved by finding an integrating factor; in this case, we see it is x^2, as (d/dx)(x^3y)=x^3y'+3x^2y=x^2(xy'+3y),
and so we have:
x^3y'+3x^2y=6x^5
(d/dx)(x^3y)=6x^5
You can integrate with respect to x on both sides (don't forget the constant of integration), and then solve for y. Then find the value of the constant such that y(1)=0, and you have your solution.
--Kevin C.