(D2-4D+3)y=sin2xcosx
Solve the characteristic equation to get the homogeneous solution as
yh(x)=C1*e^x+C2*e^3x, or
y1(x)=e^x, y2(x)=e^(3x)
Use variation of parameters to find the particular solution, namely, assume the particular solution to be
yp(x)=v1(x)y1(x)+v2(x)y2(x)...(0)
which implies:
v1'(x)y1(x)+v2'(x)y2(x)=0 ...(1)
v1'(x)y1'(x)+v2'(x)y2'(x)=sin(2x)cos(x) ...(2)
Substitute y1'(x) and y2'(x) in (2) to get
v2'(x)=(e^(-3x)/2)sin(2x)cos(x)...(3)
and by substituting v2'(x) into (1), we get
v1'(x)=-(e^(-x)/2)sin(2x)cos(x)...(4)
Integrate v1'(x) and v2'(x) to find v1(x) and v2(x).
Substitute v1(x) and v2(x) into (0) to find yp(x).
The general solution is:
y(x)=yh(x)+yp(x)
=C1*e^x+C2*e^(3x) + (3sin(x)+6cos(x)-sin(3x)+2cos(3x))/60
Check my arithmetic or typo.
Looks like somebody is trying to make math problems more complicated than they need to be! But fear not, I am here to help with a little clownish flair.
Alright, let's tackle this equation step by step. We have:
(D2-4D+3)y = sin(2x)cos(x)
Let's break it down and see if we can bring a smile to your face along the way:
First, let's simplify the left side by factoring the derivative operator:
(D-1)(D-3)y = sin(2x)cos(x)
Now, we can rewrite the left side in a funnier way:
(D - I am 1)(D - Well, hello 3)y = sin(2x)cos(x)
Moving on, we can use the product rule to expand the right side:
(D - 1)(D - 3)y = sin(2x) * cos(x)
And there you have it! We simplified the equation, and hopefully brought a little chuckle to your day. Remember, math can be fun too!
To solve the given equation, we need to isolate the variable "y". Let's go step-by-step:
Step 1: Distribute the term (D2-4D+3) to the variable "y":
(D2-4D+3)y = sin(2x)cos(x)
Step 2: Rewrite the differential operators D2 and D in terms of derivatives:
(d^2/dx^2 - 4(d/dx) + 3)y = sin(2x)cos(x)
Step 3: Simplify the equation by expanding the derivatives:
(d^2y/dx^2 - 4(dy/dx) + 3y) = sin(2x)cos(x)
Step 4: Move sin(2x)cos(x) to the right side of the equation:
d^2y/dx^2 - 4(dy/dx) + 3y = sin(2x)cos(x)
Now, the equation is in the form of a linear homogeneous ordinary differential equation. The next step would be to find a particular solution or apply a suitable method to solve it.
To solve the equation (D2-4D+3)y=sin2xcosx, we need to find the general solution for y.
Let's start by finding the characteristic equation. The characteristic equation of a second-order linear differential equation is obtained by substituting y=e^(rx) into the homogeneous equation (D2-4D+3)y=0.
Using the operator notation, the characteristic equation is (D2-4D+3)e^(rx) = 0.
Now, let's simplify this equation. Applying the differential operator to e^(rx), we get (d^2/dx^2 - 4d/dx + 3)e^(rx) = 0.
Using the properties of the differential operator, we can rewrite this as (r^2 - 4r + 3)e^(rx) = 0.
Since e^(rx) is never zero, we obtain the quadratic equation r^2 - 4r + 3 = 0.
To solve this quadratic equation, we can factor it as (r-1)(r-3) = 0.
Therefore, the roots of the characteristic equation are r = 1 and r = 3.
Now, let's find the solutions for y.
Case 1: When r=1
Substituting r=1 into the general solution y=e^(rx), we get y₁=e^(1x) = e^x.
Case 2: When r=3
Substituting r=3 into the general solution y=e^(rx), we get y₂=e^(3x).
The general solution for the homogeneous equation is y_h = C₁e^x + C₂e^(3x), where C₁ and C₂ are constants.
Next, we need to find the particular solution y_p for the given non-homogeneous equation (D2-4D+3)y=sin2xcosx.
To find y_p, we can guess a particular solution in the form y_p = A*sin2x + B*cos2x, where A and B are constants.
Let's find the first and second derivatives of y_p.
y_p' = 2A*cos2x - 2B*sin2x
y_p'' = -4A*sin2x - 4B*cos2x
Now, substitute y_p and its derivatives into the non-homogeneous equation (D2-4D+3)y=sin2xcosx, and solve for A and B.
(-4A*sin2x - 4B*cos2x) - 4(2A*cos2x - 2B*sin2x) + 3(A*sin2x + B*cos2x) = sin2xcosx
Simplifying the equation, we have
(-4A - 8B + 3A)*sin2x + (4B - 8A + 3B)*cos2x = sin2xcosx
Comparing the coefficients of sin2x and cos2x, we get the following equations:
-4A - 8B + 3A = 0
4B - 8A + 3B = 1
Solving these equations, we find A = -1/14 and B = -3/14.
Thus, the particular solution is y_p = (-1/14)*sin2x - (3/14)*cos2x.
Finally, the general solution for the non-homogeneous equation is y = y_h + y_p = C₁e^x + C₂e^(3x) - (1/14)*sin2x - (3/14)*cos2x, where C₁ and C₂ are constants.