hola chicos necesitos resolver estos ejercicios, alguien me ayudaria a realizarlos, se los agradeceria bastante.

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- Thread starter Coriche
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hola chicos necesitos resolver estos ejercicios, alguien me ayudaria a realizarlos, se los agradeceria bastante.

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Please post your "problem" in English - including your work.hola chicos necesitos resolver estos ejercicios, alguien me ayudaria a realizarlos, se los agradeceria bastante.

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Please show us what you have tried and

Please follow the rules of posting in this forum, as enunciated at:

Please share your work/thoughts about this problem

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Since I'm not going to sit here and do

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-Dan

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To find the general solution to the entire equation you just need to add a single function that satisfies the entire equation to that. Seeing that "\(\displaystyle 30e^{2x}\)" I would recommend trying a solution of the form \(\displaystyle y= Ae^{2x}\) where A is a constant you need to determine to fit this equation. Then \(\displaystyle y'= 2Ae^{2x}\) and \(\displaystyle y''= 4Ae^{4x}\). Putting those into the equation \(\displaystyle y''+3y'+ 2y= 4Ae^{4x}+ 6Ae^{2x}+ 2Ae^{2x}= 12Ae^{2x}= 30Ae^{2x}\) so \(\displaystyle A= 30/12= 5/2\). The general solution to this differential equation is \(\displaystyle y(x)= C_1e^{-2x}+ C_2e^{-x}+ (5/2)e^{2x}\).

Check; if \(\displaystyle y(x)= C_1e^{-2x}+ C_2e^{-x}+ (5/2)e^{2x}\) then \(\displaystyle y'(x)= -2C_1e^{-2x}- C_2e^{-x}+ 5e^{2x}\) and \(\displaystyle y''(x)=4C_1e^{-2x}+ C_2e^{-x}+ 10e^{2x}\).

So \(\displaystyle y''+ 3y'+ 2y= 4C_1e^{-2x}+ C_2e^{-x}+ 10e^{2x} -6C_1e^{-2x}- 3C_2e^{-x}+ 15e^{2x}+ 2C_1e^{-2x}+ 2C_2e^{-x}+ 5e^{2x}= (4C_1- 6C_1+ 2C_1)e^{-2x}+(C_2- 3C_2+ 2C_2)e^{-x}+ (10+ 15+ 5)e^{2x}= 30e^{2x}\)

Does that seem familiar to you? Whoever gave you these problem clearly expected to know it! Now that you have been reminded, try the others your self.

Queremos ayundarse, pero lo necesita que usted escriba en Inglés, para reglas del foro lo requieren. Nosotros entendemos que será defícil, pero más personas se ayudarán. Gracias por entender.

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In response#4, you have been shown how to attack these problems. Now it is your turn.Hello everyone, sorry for the previous post, but I would like you to help me only in exercises 4,16 and 17, I thank you very much for your comments, again excuse me

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Please post ONE problem in one thread. Please make separate thread for separate problem.

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As before, the associated homogeneous equation is y''+ 9y= 0 which has characteristic equation \(\displaystyle r^2+ 9= 0\). \(\displaystyle r^2= -9\), \(\displaystyle r= \pm 3i\).

That tells us that the general solution to the associated homogeneous equation is \(\displaystyle y(x)= C_1 cos(3x)+ C_2 sin(3x)\).

Now, for the "non-homogeous" part. Seeing "cos(x)" you should immediately try \(\displaystyle y(x)= A cos(x)+ B sin(x)\) where A and B are constants to be determined. \(\displaystyle y'(x)= -A sin(x)+ B cos(x)\) and \(\displaystyle y''(x)= -A cos(x)- B sin(x)\) so \(\displaystyle y''+ 9y= -A cos(x)- B sin(x)+ 9A cos(x)+ 9B sin(x)= 8A cos(x)+ 8B sin(x)= cos(x)\). 8A= 1 so \(\displaystyle A= \frac{1}{8}\), 8B= 0 so B= 0. \(\displaystyle y(x)= \frac{1}{8}cos(x)\).

For "cos(3x)" your first thought might be to try \(\displaystyle y(x)= A cos(3x)+ B sin(3x)\) but that won't work! Since cos(3x) and sin(3x) are already solutions to the associated homogeneous equation, whatever A and B are, the left side of the equation will be 0 and cannot be made equal to cos(3x).

Instead try \(\displaystyle y(x)= Ax cos(3x)+ Bx sin(3x)\). (Explaining why multiplying by x works takes a couple of pages in a Differential Equations textbook so I won't try here.) \(\displaystyle y'(x)= A cos(3x)- 3Ax sin(x)+ B sin(3x)+ 3B xcos(3x)\) and \(\displaystyle y''=-6A cos(3x)- 9Ax cos(3x)+ 6B cos(3x)- 9B xsin(3x)\). \(\displaystyle y''+ 9y= -6A cos(3x)- 9Ax cos(3x)+ 6B cos(3x)- 9B xsin(3x)+ 9A cos(3x)+ 9B sin(3x)= -6A cos(3x)+ 6B sin(3x)= cos(3x)\). \(\displaystyle -6A= 1\) so \(\displaystyle A= -\frac{1}{6}\) and \(\displaystyle 6B= 0\) so \(\displaystyle B= 0\).

The general solution to this differential equation is

\(\displaystyle y(x)= C_1 cos(3x)+ C_3 sin(3x)+ \frac{1}{8} cos(x)- \frac{1}{6} xcos(3x)\)

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For a solution to the entire equation try \(\displaystyle y(x)= (Ax^2+ Bx+ C)e^x\). Put that into the equation to see what A, B, and C must be.

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If \(\displaystyle t= ln(x)\) then \(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}\) and \(\displaystyle \frac{d^2y}{dx^2}= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{1}{x}\frac{d^2y}{dt^2}\right)= \frac{1}{x^2}\left(\frac{d^2y}{dt^2}- \frac{dy}{dt}\right)\) so that the equation becomes \(\displaystyle \frac{d^2y}{dt^2}- \frac{dy}{dt}- 2\frac{dy}{dt}+ 2y= \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= e^{3t} cos(e^t)\).