Ecuaciones diferenciales de segundo y orden superior

Coriche

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

Super Moderator
Staff member
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 exactly where you are stuck.

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

topsquark

Senior Member
hola chicos necesitos resolver estos ejercicios, alguien me ayudaria a realizarlos, se los agradeceria bastante.
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Since I'm not going to sit here and do all of them for you I will give you a hint: All of these use the method of undetermined coefficients. Can you do, say, the first one and post your work? We can look it over and spot any mistakes for you.

-Dan

HallsofIvy

Elite Member
The first one is $$\displaystyle y''+ 3y'+ 2y= 30e^{2x}$$, a non-homogeneous linear differential equation with constant coefficients. The "associated homogeneous equation" is $$\displaystyle y''+ 3y'+ 2y= 0$$ and its "characteristic equation" is $$\displaystyle r^2+ 3r+ 2= (r+2)(r+1)= 0$$. The "characteristic roots" are r= -2 and r= -1 which means that the "general solution to the associated homogeneous equation" is $$\displaystyle y_h(x)= C_1e^{-2x}+ C_2e^{-x}$$ where "$$\displaystyle C_1$$" and "$$\displaystyle C_2$$" are arbitrary constants.

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.

cbarker12

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

Coriche

New member
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

cbarker12

New member
Let's start with number 4. What is the homogenous characteristic equation for no. 4?

Subhotosh Khan

Super Moderator
Staff member
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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In response#4, you have been shown how to attack these problems. Now it is your turn.

Please show us what you have tried and exactly where you are stuck.

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

HallsofIvy

Elite Member
#4 is y''+ 9y= cos(x)+ cos(3x).

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

HallsofIvy

Elite Member
#16, $$\displaystyle y''- 4y'+ 4y= x^2e^x$$ has associated homogeneous equation $$\displaystyle y''- 4y'+ 4y= 0$$ and characteristic equation $$\displaystyle r^2- 4r+ 4= (r- 2)^2= 0$$ so has r= 2 as a double root. That means that the general solution to the associated homogeneous equation is $$\displaystyle y(x)= C_1 e^{2x}+ C_2 xe^{2x}$$.

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.

HallsofIvy

Elite Member
#17. $$\displaystyle x^2y''- 2xy'+ 2y= x^3 cos(x)$$ does NOT have "constant coefficients". It is an "Euler type" or "Equi-potential" equation (so called because the power of x in the coefficient is the same as the degree of the derivative of y). However, the substitution t= ln(x) changes it to an equation in t with constant coefficients.

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 $$\displaystyle \frac{d^2y}{dt^2}- \frac{dy}{dt}- 2\frac{dy}{dt}+ 2y= \frac{d^2y}{dt^2}- 3\frac{dy}{dt}+ 2y= e^{3t} cos(e^t)$$.​