1. ## Linear integral (still..)

How am I suppose to solve this one?
What is the way?

2. ## Convert to integral over x…

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.

3. 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.

TwistedOne151, I have a question about what you wrote here:
(d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)[(t^4-1)/4].
-5t^3+3y(t)+ty'(t)=t^3
On the left side of the first equation you have an integral with respect to x, and you differentiate it with respect to t, and you come up with the same equation, only with t as the variable, how can you do that?

10x again.

4. ## Fundamental theorem of calculus

You are familiar with the fundamental theorem of calculus, yes? I just used the form given here.
We have the derivative with respect to t of the integral ∫_{1}^{t}(-5x^3+3y+xy')dx. Now, from the fundamental theorem of calculus, if F(x) is an antiderivative of the above (that is, if F'(x)=-5x^3+3y+xy'), then the integral is
∫_{1}^{t}(-5x^3+3y+xy')dx=F(t)-F(1),
and so:
(d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)(F(t)-F(1))
=F'(t)-0
=-5t^3+3y(t)+ty'(t)

Does this explain it?

--Kevin C.

5. Originally Posted by TwistedOne151
You are familiar with the fundamental theorem of calculus, yes? I just used the form given here.
We have the derivative with respect to t of the integral ∫_{1}^{t}(-5x^3+3y+xy')dx. Now, from the fundamental theorem of calculus, if F(x) is an antiderivative of the above (that is, if F'(x)=-5x^3+3y+xy'), then the integral is
∫_{1}^{t}(-5x^3+3y+xy')dx=F(t)-F(1),
and so:
(d/dt)∫_{1}^{t}(-5x^3+3y+xy')dx=(d/dt)(F(t)-F(1))
=F'(t)-0
=-5t^3+3y(t)+ty'(t)

Does this explain it?

--Kevin C.

Yes, 10x