general solution of the given second order differential equa

general solution of the given second order differential equation y\'\'+10y\'+25y=0

Solution

For such linear homogeneous differential equations we assume solutions of the form:

y=e^{kx}

Substituting gives:

k^2+10k+25=0

(k+5)^2=0

k=-5

So it is a repeated root.

For k=-5 one solution is:e^{-5x}

For twice repeated root another solution is: xe^{-5x}

For thrice repeated root eg. (k-1)^3=0

x^2e^{-5x} would be another solution.

It is is easy to check:y2=xe^{-5x} is a solution

y2\'=e^{-5x}-5xe^{-5x}

y2\'\'=-10e^{-5x}+25xe^{-5x}

y2\'\'+10y\'+25=-10e^{-5x}+25xe^{-5x}+10(e^{-5x}-5xe^{-5x})+25xe^{-5x}=0

So general solution is:

y(x)=Ae^{5x}+Bxe^{5x}