# Diffrential Equations..

• May 2nd 2009, 12:23 PM
change_for_better
Diffrential Equations..
The set of solutions of the initial - value problem:

http://www11.0zz0.com/2009/05/02/16/649131474.jpg

for x=0, when y=1 satisfy the equation :

http://www11.0zz0.com/2009/05/02/17/571671694.jpg

I try to solve this question,but I feel my solution is false ..

http://www11.0zz0.com/2009/05/02/19/183360438.jpg
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Question 2

The particular solution of differential equation dy/dt=sin 3t

such that y=-1 at pi/3

http://www11.0zz0.com/2009/05/02/18/893356595.jpg

I select the first answer and this is my solution ..

http://www11.0zz0.com/2009/05/02/19/667636082.jpg

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Question 3
Solve the differential equation:dy/dx=f(x)
f(x)=16-x^2
subject to the condition y(3)=0

My solution:

http://www11.0zz0.com/2009/05/02/19/250526552.jpg
• May 2nd 2009, 12:33 PM
redsoxfan325
I'm going to look at the first question first.

$\frac{dx}{dy} = e^{y-x} = \frac{e^y}{e^x} \implies e^x\,dx = e^y\,dy$

Integrate and get $e^x = e^y+C$

If $(0,1)$ is satisfied, we have $1 = e+C \implies C=1-e$.

So we have $e^x=e^y+1-e \implies \boxed{e^x-e^y+e-1=0}$

The other two look fine.
• May 2nd 2009, 12:42 PM
change_for_better
Quote:

Originally Posted by redsoxfan325
I'm going to look at the first question first.

$\frac{dx}{dy} = e^{y-x} = \frac{e^y}{e^x} \implies e^x\,dx = e^y\,dy$

Integrate and get $e^x = e^y+C$

If $(0,1)$ is satisfied, we have $1 = e+C \implies C=1-e$.

So we have $e^x=e^y+1-e \implies \boxed{e^x-e^y+e-1=0}$

The other two look fine.

Thank you very much for helping me :)
• May 2nd 2009, 12:43 PM
redsoxfan325
You're welcome.
• May 2nd 2009, 03:20 PM
change_for_better

http://www9.0zz0.com/2009/05/02/22/609724238.jpg

I try to solve the question but the answer is totally diffrent from the choises

http://www2.0zz0.com/2009/05/02/22/897706459.jpg
• May 2nd 2009, 03:43 PM
change_for_better
What is The particular solution of differential equation dy/dt=sin 2t such that

y=3/2 at pi/2

Can you tell me if my solution true??

http://www2.0zz0.com/2009/05/02/22/453945533.jpg
• May 2nd 2009, 06:17 PM
redsoxfan325
Quote:

Originally Posted by change_for_better

http://www9.0zz0.com/2009/05/02/22/609724238.jpg

I try to solve the question but the answer is totally diffrent from the choises

http://www2.0zz0.com/2009/05/02/22/897706459.jpg

You're fine through this step: $\ln(e^c)=\ln(3) \implies c=\ln(3)$

But $\ln(3)\neq 1$.

So you should have $y=e^{-2x+\ln(3)} = e^{-2x}\cdot e^{\ln(3)} = \boxed{3e^{-2x}}$
• May 2nd 2009, 06:19 PM
redsoxfan325
Quote:

Originally Posted by change_for_better
What is The particular solution of differential equation dy/dt=sin 2t such that

y=3/2 at pi/2

Can you tell me if my solution true??

http://www2.0zz0.com/2009/05/02/22/453945533.jpg

This is correct.
• May 2nd 2009, 06:29 PM
change_for_better
Thank you very much redsoxfan325:)

I know my mistake now,

I calculate ln3 by calculator and the answerwill be 1.09

so this is my mistake ..