Solve the initial value problem

y′′+4y′+40y=0

y(0)= 3
y′(0)=5.

The solution is

Step-by-step explanation:

First we will write the characteristic equation, which is

This is a quadratic equation

For the general form of quadratic equation, , is given by the general formula,

or

Hence, for

and

Hence, the formula becomes

or

or

or

or

or

Hence, ±

That is, or

These are the roots of the equation.

For the general solution,

Since the roots of the equation are complex,

If the roots of a characteristic equation are in the form ± , then the general solution is given by

Hence, the general solution for the differential equation becomes

Then

Now, from the question,

y(0)= 3

y′(0)=5.

That is,

∴

[NOTE: and and ]

Also,

Now, we can find by putting the value of into the equation

∴ and

Then the solution becomes

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)