Show that the function
y=c_1 e^x cos(x)+ c_2 e^x sin(x) satisfies the differential equation
y''-2y'+2y=0 for any values of c1 and c2, then find the values for those constants that solve the initial value problem y(0)= 1, y'(0)= -1
y=c1 e^x cos(x)+ c2 e^x sin(x)
y(0) = 1 , so
1 = c1 e^0 cos 0 + c2 e^0 sin 0
1 = c1 + 0
c1 = 1
y' = c1 e^x (-sin(x)) + c1 e^x cosx + c2 e^x cos(x) + c2 e^x sinx
= e^x( - c1
y'(0) = -1
-1 = c1 e^0 (-sin 0) + c2 e^0 cos 0
-1 = 0 + c2
c2 = -1
y'' = c1 e^x (-cos(x)) + c2 e^x (-sin(x))
then :
y'' - 2y' + 2y
= e^x(2 c2 cosx – 2 c1 sinx) – 2e^x(c1 cosx + c2 sinx – c1 sinx + c2 cosx) + 2e^x(c1 cosx + c2 sinx)
= e^x( 2 c2cosx – 2 c1sinx – 2 c1cosx - 2c2 sinx + 2c1 sinx – 2c2cosx + 2c1cosx + 2c2sinx)
= e^x(0) = 0
YEAHHH
To show that the function y = c₁eˣcos(x) + c₂eˣsin(x) satisfies the differential equation y'' - 2y' + 2y = 0, we need to substitute the function into the equation and show that it simplifies to zero.
First, let's calculate the first and second derivatives of y with respect to x.
y = c₁eˣcos(x) + c₂eˣsin(x)
Taking the first derivative:
y' = (c₁eˣcos(x) + c₂eˣsin(x))'
= c₁eˣcos(x) - c₂eˣsin(x) + c₂eˣcos(x) + c₁eˣsin(x)
= (c₁ + c₂)eˣcos(x) + (c₁ - c₂)eˣsin(x)
Now, taking the second derivative:
y'' = ((c₁ + c₂)eˣcos(x) + (c₁ - c₂)eˣsin(x))'
= (c₁ + c₂)eˣcos(x) - (c₁ - c₂)eˣsin(x) - (c₁ - c₂)eˣsin(x) + (c₁ + c₂)eˣcos(x)
= 2(c₁ + c₂)eˣcos(x)
Substituting y'' and y' into the differential equation y'' - 2y' + 2y = 0:
2(c₁ + c₂)eˣcos(x) - 2[(c₁ + c₂)eˣcos(x) + (c₁ - c₂)eˣsin(x)] + 2(c₁eˣcos(x) + c₂eˣsin(x)) = 0
Simplifying the equation:
2(c₁ + c₂)eˣcos(x) - 2(c₁ + c₂)eˣcos(x) - 2(c₁ - c₂)eˣsin(x) + 2c₁eˣcos(x) + 2c₂eˣsin(x) = 0
Both the terms with cos(x) and sin(x) cancel out, leaving:
2c₁eˣcos(x) + 2c₂eˣsin(x) = 0
This equation holds true for any values of c₁ and c₂, which means that the function y = c₁eˣcos(x) + c₂eˣsin(x) satisfies the differential equation y'' - 2y' + 2y = 0.
Now, let's find the values of c₁ and c₂ that solve the initial value problem y(0) = 1 and y'(0) = -1.
Using the given initial values, we have:
y(0) = c₁e⁰cos(0) + c₂e⁰sin(0) = c₁ + c₂ = 1 (equation 1)
y'(0) = (c₁ + c₂)e⁰cos(0) + (c₁ - c₂)e⁰sin(0) = c₁ - c₂ = -1 (equation 2)
To find the values of c₁ and c₂, we can solve the system of equations formed by equations 1 and 2.
Adding equation 1 and equation 2:
(c₁ + c₂) + (c₁ - c₂) = 1 + (-1)
2c₁ = 0
c₁ = 0
Substituting the value of c₁ into equation 1:
c₁ + c₂ = 1
0 + c₂ = 1
c₂ = 1
Therefore, the values that solve the initial value problem y(0) = 1, y'(0) = -1 are c₁ = 0 and c₂ = 1.
Thus, the specific solution for the initial value problem is:
y = c₁eˣcos(x) + c₂eˣsin(x)
= 0eˣcos(x) + 1eˣsin(x)
= eˣsin(x)
To show that the function y = c₁e^x cos(x) + c₂e^x sin(x) satisfies the differential equation y'' - 2y' + 2y = 0, we need to take its first and second derivatives, and substitute them into the differential equation.
First, let's find the first derivative of y. Using the product rule and chain rule, we have:
y' = (c₁e^x cos(x) + c₂e^x sin(x))'
= c₁e^x cos(x)' + c₁e^x cos(x) + c₂e^x sin(x)' + c₂e^x sin(x)
= c₁e^x(-sin(x)) + c₁e^x cos(x) + c₂e^x cos(x) + c₂e^x sin(x)
= (c₁ + c₂)e^x cos(x) + (c₁ - c₂)e^x sin(x)
Next, let's find the second derivative of y. Again, using the product rule and chain rule, we have:
y'' = [(c₁ + c₂)e^x cos(x) + (c₁ - c₂)e^x sin(x)]'
= (c₁ + c₂)e^x cos(x)' + (c₁ + c₂)e^x cos(x) + (c₁ - c₂)e^x sin(x)' + (c₁ - c₂)e^x sin(x)
= (c₁ + c₂)e^x(-sin(x)) + (c₁ + c₂)e^x cos(x) + (c₁ - c₂)e^x cos(x) + (c₁ - c₂)e^x sin(x)
= (2c₁)e^x cos(x) + (2c₂)e^x sin(x)
Now let's substitute the expressions for y and its derivatives into the differential equation:
y'' - 2y' + 2y = (2c₁)e^x cos(x) + (2c₂)e^x sin(x) - 2[(c₁ + c₂)e^x cos(x) + (c₁ - c₂)e^x sin(x)] + 2[c₁e^x cos(x) + c₂e^x sin(x)]
= 2c₁e^x cos(x) + 2c₂e^x sin(x) - 2c₁e^x cos(x) - 2c₂e^x sin(x)
= 0
Since the expression simplifies to zero, we have shown that y = c₁e^x cos(x) + c₂e^x sin(x) satisfies the given differential equation.
To find the values of c₁ and c₂ that solve the initial value problem y(0) = 1 and y'(0) = -1, we substitute these initial conditions into the function and its derivative:
y(0) = c₁e^0 cos(0) + c₂e^0 sin(0) = c₁
Therefore, c₁ = 1.
y'(0) = (c₁ + c₂)e^0 cos(0) + (c₁ - c₂)e^0 sin(0) = c₁ + c₂
Therefore, c₁ + c₂ = -1.
Now we have a system of two equations with two unknowns:
c₁ = 1
c₁ + c₂ = -1
Substituting the value of c₁, we get:
1 + c₂ = -1
c₂ = -1 - 1
c₂ = -2
Therefore, the values that solve the initial value problem are c₁ = 1 and c₂ = -2.