Given y 10y 25y 3e5x a Find ynx b Find ypxSolutiona yhx i

Given y\" - 10y\' + 25y = 3e^5x a.) Find y_n(x) b.) Find y_p(x)

Solution

a)

yh(x) is solution to the homogeneous part of the ODE ie: y\'\'-10y\'+25y=0

Let, exp(kx) be a solution

Substituting gives

k^2-10k+25=0

k=5 ie repeated roots

So, yh(x)=e^{5x}(A+Bx)

b)

yp is particular solution ie some function which satisfies the given ODE.

Based on inhomogeneous part we can guess what is the solution

Here it is :3 e^{5x}

SO our guess would be Ce^{5x}

But in this case e^{5x} is already solution to the homogeneous ODE so the guess then becomes

Cxe^{5x}

But then xe^{5x} is also solution to the ODE. So the guess becomes

yp=Cx^2e^{5x}

yp\'=2Cxe^{5x}+5Cx^2e^{5x}

yp\'\'=2Ce^{5x}+20Cxe^{5x}+25Cx^2e^{5x}

Substituting gives

2Ce^{5x}+20Cxe^{5x}+25Cx^2e^{5x}-10(2Cxe^{5x}+5Cx^2e^{5x})+25(Cx^2e^{5x})=3e^{5x}

2Ce^{5x}=3e^{5x}

Hence, C=3/2

So,yp=3/2 x^2e^{5x}