#### anetajakimova

##### New member

- Joined
- Jun 1, 2019

- Messages
- 6

- Thread starter anetajakimova
- Start date

- Joined
- Jun 1, 2019

- Messages
- 6

- Joined
- Jun 18, 2007

- Messages
- 20,882

Since your "forcing function" is also "homogeneous solution", you have a "resonance effect".View attachment 12399View attachment 12400View attachment 12401

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

Thank you.

The total solution should be:

y = A*cos(2x) + B*sin(2x) + x * [cos(2x) + sin(2x)]

Check that above function satisfies the given DE.

Now using initial conditions, solve for A and B.

- Joined
- Nov 24, 2012

- Messages
- 2,909

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.

- Joined
- Jun 1, 2019

- Messages
- 6

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

- Joined
- Jun 1, 2019

- Messages
- 6

Okay no problem,where ?

- Joined
- Jun 1, 2019

- Messages
- 6

- Joined
- Jun 1, 2019

- Messages
- 6

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.

- Joined
- Nov 24, 2012

- Messages
- 2,909

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

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.

- Joined
- Jan 27, 2012

- Messages
- 6,022

I am puzzled by this. You are trying to do differential equations but do not know how to differentiate?

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.

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)).