Question

(10 points) (a) Find the general solution to y′′−6y′+9y=0y′′−6y′+9y=0. Enter your answer as y=…y=… . In your answer, use c1c1 and c2c2 to denote arbitrary constants and xx the independent variable. Enter c1c1 as c1 and c2c2 as c2. help (equations) (b) Find the solution that satisfies the initial conditions y(0)=5y(0)=5 and y′(0)=0y′(0)=0. help (equations)

Answer 1

a) The general solution of the differential equation is
`y(t)=c_1e^(3t)+c_2te^(3t)`.

b) Applying the initial conditions the solution to the differential equation is then
`y(t)=5e^(3t)-15e^(3t)t`.

Explanation:

We have the following second order linear homogeneous differential equation with constant coefficients
`y''-6y'+9y=0`

Given the differential equation
`ay'' + by' + cy = 0`,
`a\\eq 0`, consider the quadratic polynomial
`ax^2+bx+c`, called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them
`r` and
`s`. The general solution of the differential equation is:

a)
`\ds y=Ae^(rt)+Be^(st)`, if the roots
`r` and
`s` are real numbers and
`r\\eq s`.

b)
`\ds y=Ae^(rt)+Bte^(rt)`, if
`r=s` is real.

c)
`\ds y=A\cos(\beta t)e^(\alpha t)+B\sin(\beta t)e^(\alpha t)`, if the roots
`r` and
`s` are complex numbers
`\alpha+\beta i` and
`\alpha-\beta i`.

Applying the above information we have that:

The characteristic polynomial is

`x^2-6x+9=0`

Its roots are

`x^2-6x+9=(x-3)^2=0\\x=3`

So, the general solution of the differential equation is:

`y(t)=c_1e^(3t)+c_2te^(3t)`

and its derivative is:

`y'(t)=(d)/(dt)\left(c_1e^(3t)+c_2te^(3t)\right)=c_1e^(3t)\cdot \:3+c_2\left(e^(3t)+3e^(3t)t\right)`

Now, plug in the initial conditions to get the following system of equations.

`y(0)=c_1e^(3\cdot 0)+c_2\cdot 0\cdot e^(3\cdot 0)=5\\\\y'(0)=c_1e^(3\cdot 0)\cdot \:3+c_2\left(e^(3\cdot 0)+3e^(3\cdot 0)\cdot 0\right)=0`

Solving this system gives

`c_1=5` and
`c_2=-15`

The actual solution to the differential equation is then
`y(t)=5e^(3t)-15e^(3t)t`