Question

In Problems 37-40 solve the given initial-value problem.

y^''-10y+25y=0

y(0)=1, y(1)=0

Short answer

y=e^5x-xe^5e

Step-by-step solution

Determine the initial value of differential equation

Given,

y^''-10y^'+25y=0

Initial conditions:

y(0)=1, y(1)=0

Consider y=e^mx as the solution of the differential equation,

Substitute y=e^mx, y^'=me^mx and y^"=m²e^mxinto y^''-10y^'+25y=0 to obtain the auxiliary equation.

m²e^mx-10me^mx+25e^mx=0

e^mt(m²-10m+25)=0

For any real x, e^mx≠ 0, we have

Then the roots are

m_1,2=5

Find the additional linearly independent solution

The need to have an additional linearly independent solution for this repeated root, can obtain the general solution of our differential equation .

y^''-10y^'+25y=0

y=ne^x(k+sx)

y=c₁e^x+c₂xe^x----1

Real and repeated

Finally, since we have repeated root m_1,2=5 , then we need to have an additional linearly independent solution for this repeated root, then we obtain the general solution of our differential equation.

y=ne^5x(k+sx)

y=c₁e^5x+c₂xe^5x

Arbitary constant:

c₁=n×k, c₂=n×s

Now we have to apply the given conditions in the general solution of our given differential equations as the following technique :

We have to apply with the point (x,y)=(0,1) in the general solution as

1=c₁e^o+c₂(0)

c₁=1

After that, we have to apply with the point (x,y)=(1,0) in the general solution as by solving the two equations

0=c₁e⁵+c₂e⁵

c₂=-c₁

c₁=1 and c₂=-1

Substitute with the values of c₁ and c₂ into the general solution of our given differential equation.

y=e^5x-xe^5e

The solution of the differential equation isy=e^5x-xe^5e

Hence, the general solution is y=e^5x-xe^5e