# Thread: 2nd order DE's with constant coefficients.

1. ## 2nd order DE's with constant coefficients.

1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
e^(3x) and xe^(3x)

Thanks for your help.

2. What ideas have you had so far?

3. Originally Posted by hejmh
1- How can I solve this equation? y''+4y'=0

2-What is the diffrential equation of type y''+ay'+by=0 which has the solution of:
e^(3x) and xe^(3x)

Thanks for your help.
$\displaystyle y''+4y'=0\Rightarrow m^2+4m=0$

Do you know how to use this form?

4. For part 2, you're told $\displaystyle \displaystyle y = e^{3x}$ is a solution.

So $\displaystyle \displaystyle y' = 3e^{3x}$ and $\displaystyle \displaystyle y'' = 9e^{3x}$.

Substitute into the DE to get $\displaystyle \displaystyle 9e^{3x} + 3a\,e^{3x} + b\,e^{3x} = 0$.

Now do the same with the other solution $\displaystyle \displaystyle y = x\,e^{3x}$ to get another equation that involves $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$. Then you can solve the two equations simultaneously for $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$.

5. Originally Posted by dwsmith
$\displaystyle y''+4y'=0\Rightarrow m^2+4m=0$

Do you know how to use this form?
From this point, you factor and solve for m.

$\displaystyle m(m+4)=0\Rightarrow m=0,-4$

$\displaystyle y=C_1e^{m_1x}+C_2e^{m_2x}$

$\displaystyle y=C_1e^{0*x}+C_2e^{-4x}=C_1+C_2e^{-4x}$

6. Yes. thanks

7. Originally Posted by hejmh
1- How can I solve this equation? y''+4y'=0
Here's another way to do this. Since there is no "y" (undifferentiated) let u= y' so the equation becomes u'+ 4u= 0. That is the same as du/dx= -4u which is a separable, first order differential equation. You can write it as du/u= -4 dx and integrate both sides: ln(u)= -4x+ c which leads to $\displaystyle u(x)= y'(x)= e^{-4x+ c}= Ce^{-4x}$ where $\displaystyle C= e^{c}$. Now find y by integrating $\displaystyle y'= Ce^{-4x}$: $\displaystyle y(x)= \frac{C}{-4}e^{4x}+ B$ which we can also write as $\displaystyle y(x)= Ae^{-4x}+ B$ where now $\displaystyle A= \frac{C}{-4}= \frac{e^c}{-4}$.