The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation. y'' − 3y' + 2y = 7e3x; y1 = ex

Step-by-step explanation

Standard form

Hence your P(t) = -3, Q(t) = 2

After replacing all y", y' and y to homogeneous part, you will have

because

Let U = V'

.

Replace back,

then

So, general solution of the ODE is

Particular solution is just take derivative of the general one twice and plug back into the original ODE to find A and B

You can finish it by yourself. Let me know if you need more help

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation: