Solving the IVP and minimum value

[DrunkApple]

Homework Statement

Solve the initial value problem
y' = 2cos (2x)/(3+2y), y(0) = 1
and determine where the solution attains its maximum value

Homework Equations

The Attempt at a Solution

I got it to here
y^{2} + 3y = sin(2x) + C
but I don't know what to do from here...
Help please

 

[sunjin09]

Homework Statement

Solve the initial value problem
y' = 2cos (2x)/(3+2y), y(0) = 1
and determine where the solution attains its maximum value

Homework Equations

The Attempt at a Solution

I got it to here
y^{2} + 3y = sin(2x) + C
but I don't know what to do from here...
Help please

What you got there is wrong, the standard solution is write y'=dy/dx and put all y's on left and all x's on right, then integrate both sides.

 

[DrunkApple]

I did.
dy/dx = 2cos (2x)/(3+2y)
(3+2y)dy = 2cos(2x)dx
isn't this not it?

 

[sunjin09]

I did.
dy/dx = 2cos (2x)/(3+2y)
(3+2y)dy = 2cos(2x)dx
isn't this not it?

Sorry my bad, I miss-read your equation :) I guess then y = solve the 2nd order algebraic equation, which has two solutions, only one is feasible for y[0]=1; Actually set x=0 and y=1, you figure C=4, then you realize there are 2 solutions, only one gives you back y[0]=1.

BTW I was wondering why 1st order ODE has two independent solutions...

 

[DrunkApple]

would you please show me how to do it? because I really have no idea about this. How do I solve for the 2nd order?

 

[QuarkCharmer]

I did.
dy/dx = 2cos (2x)/(3+2y)
(3+2y)dy = 2cos(2x)dx
isn't this not it?

Separate the variables and integrate just like you were doing here, you were on the right track:
(3+2y)dy = 2cos(2x)dx
\int(3+2y)dy = \int2cos(2x)dx

You'll end up with a function of x's and y's, to which you can apply your initial condition and solve for your constant of integration. Once you have that, you're basically done. From there, you can simply maximize the function to find x where y_max.

 

[sunjin09]

would you please show me how to do it? because I really have no idea about this. How do I solve for the 2nd order?

solution to ax^2+bx+c=0 is given by x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

 

[DrunkApple]

so I use quadratic equation after integrating?

3y + 2y^{2} = sin (2x) + C ??

 

[Dick]

so I use quadratic equation after integrating?

3y + 2y^{2} = sin (2x) + C ??

Yes, find the solution of the ode and solve for y. Then the extreme values are where y'(x)=0. You can actually read those values of x out of the original ode.

 

[DrunkApple]

ohhh ok i got it ;D
would you give me some hints about how to get its maximum value (for 0 ≤ x ≤ 0.309)?

 

[Dick]

ohhh ok i got it ;D
would you give me some hints about how to get its maximum value (for 0 ≤ x ≤ 0.309)?

Did you work out what y(x) is? Did you figure out where the critical points where y'(x)=0 are? If you are looking for the max on [0,0.309] it will be one of the end points. Why?