1. ## Diffrential Equations..

The set of solutions of the initial - value problem:

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

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

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Question 2

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

such that y=-1 at pi/3

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

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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:

2. 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.

3. 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

4. You're welcome.

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

6. 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??

7. Originally Posted by change_for_better

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

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

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

So you should have $y=e^{-2x+\ln(3)} = e^{-2x}\cdot e^{\ln(3)} = \boxed{3e^{-2x}}$

8. 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??

This is correct.

9. 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 ..