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.