# Thread: second order odes

1. ## second order odes

hey i'm really stuck on these two equations. Can someone please please help me?

y''-4y'+3y = 8exp(-x)- 2exp(x) y(0)= 1 y'(0)= 0

x*2 y'' +xy' + y = sin(ln(x))

thanks so much!x

2. It will take some work, bu here follow the solution for the first problem. We can rewrite the DE like follows

$\displaystyle r^{2}-4r+3=0$ we firstly consider the homogenous part
$\displaystyle (r-3)(r-1)=0$ which has roots at r=3 and r=1

so we know the solution to the homogenous DE and can write it as follows

$\displaystyle y_{hom}=c_{1}e^{3x}+c_{2}e^{x}$

now we will take the right hand side into account. Since the right hand side of the DE includes one of the terms we find in the homegenous solutions we need to regard the following terms

$\displaystyle y_{try}=Ae^{-x}+Bx^{2}e^{x}+Cxe^{x}$

if you now plug this into the DE and group the terms with the same x you can determine the constants A, B and C. I found them to be respectively 1, 0 and 1. Then finally for the general solution you just combine everything in the following way

$\displaystyle y_{general}=c_{1}e^{3x}+c_{2}e^{x}+Ae^{-x}+Bx^{2}e^{x}+Cxe^{x}$

and use the initial conditions to solve for the remaining constants

The second problem will require some more work but you could attack it in the same way by first finding the solution of the homogenous DE and afterward incorporating the right hand side.

3. Originally Posted by CarrieB
hey i'm really stuck on these two equations. Can someone please please help me?

x*2 y'' +xy' + y = sin(ln(x))

thanks so much!x
The solution is:

$\displaystyle y(x)=C_1 \cos(\ln x) + C_2 \sin(\ln x) - \frac{1}{2} \ln x \cos(\ln x)$

4. A confortable procedure to solve the DE...

$\displaystyle x^{2}\cdot y^{''} + x\cdot y^{'} + y = \sin \ln x$ (1)

... can require a change of variables as illustrated in...

http://www.mathhelpforum.com/math-he...-question.html

The solution seems to be ...

$\displaystyle y= c_{1}\cdot \cos \ln x + c_{2} \cdot \sin \ln x + \frac{1}{2}\cdot \ln x \cdot \cos \ln x$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. That second equation is an "Euler type" or "equipotential" equation (the "power" of x in each term is equal to the order of the derivative) and, as ChiSigma shows, the substitution u= ln(x) changes it to an equation with constant coefficients. That can then be solved by finding its characteristic equation.

6. Originally Posted by chisigma
A confortable procedure to solve the DE...

$\displaystyle x^{2}\cdot y^{''} + x\cdot y^{'} + y = \sin \ln x$ (1)

... can require a change of variables as illustrated in...

http://www.mathhelpforum.com/math-he...-question.html

The solution seems to be ...

$\displaystyle y= c_{1}\cdot \cos \ln x + c_{2} \cdot \sin \ln x + \frac{1}{2}\cdot \ln x \cdot \cos \ln x$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Not quite, but almost right. See my previous post.