(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)

a) The general solution of the differential equation is .

b) Applying the initial conditions the solution to the differential equation is then .

Step-by-step explanation:

We have the following second order linear homogeneous differential equation with constant coefficients

Given the differential equation , , consider the quadratic polynomial , called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them and . The general solution of the differential equation is:

a) ,  if the roots and are real numbers and .

b) , if is real.

c) ,  if the roots and are complex numbers and .

Applying the above information we have that:

The characteristic polynomial is

Its roots are

So, the general solution of the differential equation is:

and its derivative is:

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

Solving this system gives

and

The actual solution to the differential equation is then

angles 1 and 3 are vertical angles, meaning that they are equal to each other. to solve this you may set angle 1 and 3 equal to each other and just solve for x like

110 = 2x + 10

100 = 2x

50 = x

check:

110 = 2* 50 + 10

110 = 100 + 10

110 = 110

hope this !

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