Question

(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13

Answer 1

Answer:

$y(x)=3e^{5x}-2xe^{5x}$

Step-by-step explanation:

The given differential equation is $y''-10y'+25y=0$

The characteristics equation is given by

$r^2-10r+25=0$

Finding the values of r

$r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5$

We got a repeated roots. Hence, the solution of the differential equation is given by

$y(x)=c_1e^{5x}+c_2xe^{5x}...(i)$

On differentiating, we get

$y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)$

Apply the initial condition y (0)= 3 in equation (i)

$3=c_1e^{0}+0\\\\c_1=3$

Now, apply the initial condition y' (0)= 13 in equation (ii)

$13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2$

Therefore, the solution of the differential equation is

$y(x)=3e^{5x}-2xe^{5x}$