y" +4y=4cos(2x) , y(0)=0 ; y'(0)=0

Hello, and welcome to FMH! :)

I agree the the homogeneous solution is:

[MATH]y_h=c_1\cos(2x)+c_2\sin(2x)[/MATH]
And so this means our particular solution will take the form:

[MATH]y_p(x)=x(A\sin(2x)+B\cos*(2x))[/MATH]
And so using the method of undetermined coefficients, we get:

[MATH]4(A-Bx)\cos(2x)-4(Ax+B)\sin(2x)+4(x(A\sin(2x)+B\cos(2x)))=4\cos(2x)[/MATH]
[MATH]A\cos(2x)-B\sin(2x)=1\cdot\cos(2x)+0\cdot\sin(2x)[/MATH]
Hence:

[MATH](A,B)=(1,0)[/MATH]
And so:

[MATH]y_p(x)=x\sin(2x)[/MATH]
And thus, the general solution is:

[MATH]y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+x\sin(2x)[/MATH]
Now, you may proceed to determine the values of the two parameters which satisfy the IVP. :)
 
Hi
I understand everything but i have just problems with derivating.
Can you explain me in details please,because i make mistake in yp when it needs to be derivated and then when you need to put it in the given equation : yp" +4yp=4cos(2x)
Thank you
 
Your posted work is too small for me to comfortably read...can you post your work where I can easily read it?
 
Okay,is it okay like this?
 

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Aren't those exactly the same small hard to read images you initially posted?
 
Unfortunately, 2 of the images are sideways, and the third is upside down. Please just type out your attempt to differentiate the particular solution so I can easily read it, and try to see where you may have gone wrong. :)
 
Okay.
I make mistake when i need to derivate this part:
yp=x (A*cos(2x) +B*sin(2x) )
i dont know how to make first and second derivative.
 
anetajakimova said:
Okay.
I make mistake when i need to derivate this part:
yp=x (A*cos(2x) +B*sin(2x) )
i dont know how to make first and second derivative.
Click to expand...

Can you post your work, using \(\LaTeX\) or even plain text so I can see where you're going wrong?
 
anetajakimova said:
Okay.
I make mistake when i need to derivate this part:
yp=x (A*cos(2x) +B*sin(2x) )
i dont know how to make first and second derivative.
Click to expand...
I am puzzled by this. You are trying to do differential equations but do not know how to differentiate?

Hopefully you know that the derivative of cos(x) is -sin(x) and the derivative of sin(x) is cos(x) and, using the "chain rule" the derivative of cos(2x) is -sin(2x) times the derivative of 2x so -2sin(2x) and, similarly, the derivative of sin(2x) is 2cos(2x). By the product rule, the derivative of x(A cos(2x)+ B sin(2x)) is (the derivative of x) times (A cos(2x)+ B sin(2x)) plus x times the derivative of A cos(2x)+ B sin(2x). That is, the derivative of x(A cos(2x)+ B sin(2x)) is (A cos(2x)+ B sin(2x)+ x(-2A sin(2x)+ 2B cos(2x)).
 
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