Homogeneous Linear Differential Equation

[engineertobe]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

 

[Deleted member 4993]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

As I understand it - it is pretty straight forward:

y' = (6)[ae^6x] + (-3)[bsin(3x)] + (3)[c cos(3x)] ..... and so on...

 

[afrazer721]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

Second order homegenoeous differential equations solutions can be found by examining polynominals, called auxiliary equations.

 

[Deleted member 4993]

Second order homegenoeous differential equations solutions can be found by examining polynominals, called auxiliary equations.

s/he is not looking for solution to DE - the solution is given to the student and the student has to formulate the homogeneous DE.

 

[afrazer721]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

Let y^(3)=0
Homogeneous equations solutions can be found by related polynomials, called auxiliary equations. In this case the auxiliary equation in my estimate is A^2+A+1=0 which illustrates the general pattern such as a recession from an economic viewpoint. This can be solved using the quadratic formula to get
A=d1=(-1+3^1/2*i)/2 and A=d2=(-1-3^1/2*i)/2. The theory of homogeneous equations then tells us that
y=e^d1*x and y=e^d2*x are solutions to the original equation, and the general solution is
y=Be^r1*x+Ce^r2*x (for any constants B and C).

 

[afrazer721]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

I got an email pertaining to the DE. It seems you want to get back to the original equation. You've finished with the solution y(x)=ae^(6x) + bcos(3x)+ csin(3x). Differentiate both sides of the equation. For example, lets say I finished with the solution y(x)= x^3+4x^2-7x-sinx+c. When I differentiate both sides, I obtain dy/dx=3x^2+8x-7-cosx. Adding cosx to both sides produces the ODE I began with:dy/dx+cosx=3x^2+8x-7. I hope this helps.

 

[afrazer721]

I am given y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I am then asked to convert it into the form y^(3)+ __y" + ___y'+ __y=0 solving for the constants. I don't know how to deal with the sin and cos. Any ideas?

I'm assuming this is an elementary second-order differential equation. We ended with the solution
y(x)=ae^(6x) + bcos(3x)+ csin(3x) where a, b, and c are constants. I'll be using the product and/or chain rule(s) to differentiate terms to get back to the second-order differential equation.

Differentiating both sides of the solution gives us
y'=dy/dx=6ae^(6x)-3bsin(3x)+3ccos(3x)
Differentiating both sides of the solution again yields
y"=d^2y/dx^2=36ae^(6x)-9bcos(3x)-9csin(3x)
The solution equation is y=ae^(6x) + bcos(3x)+ csin(3x)
Now plug y, y', and y" into y^(3)+_y"+_y'+_y=0, respectively. I'm not sure about what's going on with the value for y^(3).