# Verifying solutions for differential equations

• Jan 18th 2011, 12:49 PM
duaneg37
Verifying solutions for differential equations
Verify this is a solution of this differential equation:

S=integrand......integrate from 0 to x
dy/dx+2xy = 1 ; y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

[Difficulty]
I'm not sure how to simplify y. It looks like the fundamental theorem
of calculus. So e^(-x^2)S(from 0 to x)e^(t^2)dt should just be 1,
right? I let y=1+c[1]e^(-x^2) and used implicit differentiation, but
I end up with dy/dx+2xy=2x. If 2x=1 then I'd have it. I also
substituted y and dy/dx into the equation, but I get the same thing:
2x=1. I have to be simplifying y wrong. Can you help me? Thanks!

[Thoughts]
y=1+ce^(-x^2)
(y-1)/e^(-x^2)=c
e^(x^2)y-e^(x^2)=c
e^(x^2)y'+2xye^(x^2)-2xe^(x^2)=0
e^(x^2)y'=2xe^(x^2)-2xye^(x^2)
y'=2x-2xy
y'+2xy=2x

dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)
-2xce^(-x^2)+2x(1+ce^(-x^2))=1
-2xce^(-x^2)+2x+2xce^(-x^2)=1
2x=1
• Jan 18th 2011, 12:52 PM
dwsmith
Quote:

Originally Posted by duaneg37
Verify this is a solution of this differential equation:

S=integrand......integrate from 0 to x
dy/dx+2xy = 1 ; y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

[Difficulty]
I'm not sure how to simplify y. It looks like the fundamental theorem
of calculus. So e^(-x^2)S(from 0 to x)e^(t^2)dt should just be 1,
right? I let y=1+c[1]e^(-x^2) and used implicit differentiation, but
I end up with dy/dx+2xy=2x. If 2x=1 then I'd have it. I also
substituted y and dy/dx into the equation, but I get the same thing:
2x=1. I have to be simplifying y wrong. Can you help me? Thanks!

[Thoughts]
y=1+ce^(-x^2)
(y-1)/e^(-x^2)=c
e^(x^2)y-e^(x^2)=c
e^(x^2)y'+2xye^(x^2)-2xe^(x^2)=0
e^(x^2)y'=2xe^(x^2)-2xye^(x^2)
y'=2x-2xy
y'+2xy=2x

dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)
-2xce^(-x^2)+2x(1+ce^(-x^2))=1
-2xce^(-x^2)+2x+2xce^(-x^2)=1
2x=1

That is difficult to want to go through since it isn't in Latex and is crammed together.

I would use an integrating factor.

$\displaystyle \displaystyle\int\left\frac{d}{dx}\left(\exp{\left (\int P(x)dx\right)}y\right)\right]dx=\int\exp{\left(\int P(x)dx\right)}Q(x)dx$

$\displaystyle P(x) = 2x$

$\displaystyle Q(x) = 1$

What is this y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

Another problem?
• Jan 18th 2011, 01:02 PM
pickslides
Quote:

Originally Posted by dwsmith
What is this y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

Another problem?

This is the solution, the OP needs the derivative of this function to check if its a solution of the DE given above.
• Jan 18th 2011, 01:09 PM
duaneg37
Is it y' = -2xce^(-x^2) ? How can I get Latex? What is an OP?
• Jan 18th 2011, 01:17 PM
pickslides
Quote:

Originally Posted by duaneg37
Is it y' = -2xce^(-x^2) ?

Maybe does it satisfy $\displaystyle \frac{dy}{dx}+2xy = 1$

Quote:

Originally Posted by duaneg37
How can I get Latex?

Use MATH tags [tex]\frac{dy}{dx}+2xy = 1[/ math] (remove space between "/ math") gives $\displaystyle \frac{dy}{dx}+2xy = 1$

Double click on the equation for more info.

Quote:

Originally Posted by duaneg37
What is an OP?

Original Post.
• Jan 18th 2011, 01:18 PM
Ackbeet
You can see how to use LaTeX by looking here. OP stands for "Original Post".

Your problem: verify the solution

$\displaystyle \displaystyle y = e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}}$ for the DE

$\displaystyle y'+2xy=1.$ Correct?

You'll need to use the product rule, the chain rule, and the Fundamental Theorem of the Calculus to do this verification:

$\displaystyle \displaystyle y'=\left(e^{-x^{2}}\right)\frac{d}{dx}\left(\int_{0}^{x} e^{t^{2}}\, dt\right)+\left(-2xe^{-x^{2}}\right)\left(\int_{0}^{x} e^{t^{2}}\, dt\right)-2Cxe^{-x^{2}}.$

You'll need to finish this computation, and then plug the result into the DE on the LHS, and see if you get the required 1 to equal the RHS. Make sense?

[EDIT] Sorry, didn't see pickslides' post.
• Jan 18th 2011, 02:27 PM
duaneg37
$\displaystyle y'+2xy=1$

$\displaystyle \displaystyle y = e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}}$ Verify this is a solution of the differential equation above

$\displaystyle \displaystyle y'=\left(e^{-x^{2}}\right)\frac{d}{dx}\left(\int_{0}^{x} e^{t^{2}}\, dt\right)+\left(-2xe^{-x^{2}}\right)\left(\int_{0}^{x} e^{t^{2}}\, dt\right)-2Cxe^{-x^{2}}$

$\displaystyle \displaystyle y'=1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}$

$\displaystyle \displaystyle 1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}+2x\left( e^{-x^{2}}\int_{0}^{x} e^{t^{2}}\, dt +C\,e^{-x^{2}}\right)=1$

$\displaystyle \displaystyle 1-2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt-2Cxe^{-x^{2}}+2xe^{-x^{2}}\int_{0}^{x}e^{t^{2}}\, dt+2Cxe^{-x^{2}}=1$

$\displaystyle \displaystyle 1=1$

I think I've got it now.....Thanks for all your help!
• Jan 18th 2011, 03:03 PM
Ackbeet
I'm afraid it's incorrect. You can't just get rid of integrals like that. You did correctly use the FTC to simplify the first term, but the second term cannot be simplified. I think you'll find it cancels with terms in the 2xy of the original DE, as does the last term with the constant. Just leave the integrals there!