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.
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.
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 $\displaystyle \displaystyle b$.
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 $\displaystyle A= \frac{C}{-4}= \frac{e^c}{-4}$.