1. ## Differenntial equation

Who can tell me what's the Deffrential equation of of this general equation y=(Ae^5x)+(Bxe^5x)

2. ## Differenntial equation

Hi Hama

Please note that if a differential equation has a repeated root of characteristics equation
${(r - \alpha)^2} = 0$
then the solution will be
$y = A{e^{\alpha x}} + Bx{e^{\alpha x}}$

3. ## Re: Differenntial equation

Another way (if you do not know about the characteristic equation):

from $y= Ae^{5x}+ Bxe^{5x}$, $y'= 5Ae^{5x}+ Be^{5x}+ 5Bxe^{5x}= (5A+ B)e^{5x}+ 5Bxe^{5x}$ and $y''= 5(5A+ B)e^{5x}+ 5Be^{5x}+ 25Bxe^{5x}= (25A+ 10B)e^{5x}+ 25Be^{5x}$.

Subtracting 10 times the second equation from the third, $y''- 10y'= (25A+ 10B)e^{5x}+ 25Be^{5x}- (50A+ 10B)e^{5x}- 50Bxe^{5x}= -(25Ae^{5x}+ 25Bxe^{5x})$. Adding 25y eliminates the arbitrary constants, A and B, and gives $y''- 10y'+ 25y= 0$.

4. ## Re: Differenntial equation

Originally Posted by Hama
Who can tell me what's the Deffrential equation of of this general equation y=(Ae^5x)+(Bxe^5x)