What ideas have you had so far?
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.
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 where . Now find y by integrating : which we can also write as where now .