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

anetajakimova

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I solved till there,and than? Can someone help me?
Thank you.
 

Subhotosh Khan

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MarkFL

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Hello, and welcome to FMH! :)

I agree the the homogeneous solution is:

\(\displaystyle y_h=c_1\cos(2x)+c_2\sin(2x)\)

And so this means our particular solution will take the form:

\(\displaystyle y_p(x)=x(A\sin(2x)+B\cos*(2x))\)

And so using the method of undetermined coefficients, we get:

\(\displaystyle 4(A-Bx)\cos(2x)-4(Ax+B)\sin(2x)+4(x(A\sin(2x)+B\cos(2x)))=4\cos(2x)\)

\(\displaystyle A\cos(2x)-B\sin(2x)=1\cdot\cos(2x)+0\cdot\sin(2x)\)

Hence:

\(\displaystyle (A,B)=(1,0)\)

And so:

\(\displaystyle y_p(x)=x\sin(2x)\)

And thus, the general solution is:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(2x)+c_2\sin(2x)+x\sin(2x)\)

Now, you may proceed to determine the values of the two parameters which satisfy the IVP. :)
 

anetajakimova

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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
 

MarkFL

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Your posted work is too small for me to comfortably read...can you post your work where I can easily read it?
 

anetajakimova

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Okay no problem,where ?
 

MarkFL

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Preferably here in this thread.
 

anetajakimova

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Okay,is it okay like this?
 

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MarkFL

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Aren't those exactly the same small hard to read images you initially posted?
 

anetajakimova

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Mistake, this is what i wanted to send.
 

MarkFL

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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. :)
 

anetajakimova

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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.
 

MarkFL

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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.
Can you post your work, using \(\LaTeX\) or even plain text so I can see where you're going wrong?
 

HallsofIvy

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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.
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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