Find the particular solution to y"=3sin(x) given the general solution
y=-3sin(x) +Ax+B and the initial conditions y (π/2)=π and y'(π/2)=2
same old same old ...
y" = 3sinx
y' = -3cosx+A
since y'(π/2) = 2,
y' = -3cosx+2
y = -3sinx+2x+B
since y(π/2) = π, -3+π+B = π
y = -3sinx+2x+3
To find the particular solution to the differential equation y" = 3sin(x) with the given initial conditions, you can follow these steps:
Step 1: Find the first derivative of the general solution y = -3sin(x) + Ax + B to obtain y'.
y' = -3cos(x) + A
Step 2: Find the second derivative of the general solution y = -3sin(x) + Ax + B to obtain y''.
y'' = 3sin(x)
Step 3: Apply the initial condition y(π/2) = π.
Substitute x = π/2 and y = π into the general solution y = -3sin(x) + Ax + B.
π = -3sin(π/2) + A(π/2) + B
π = -3 + (Aπ)/2 + B (since sin(π/2) = 1)
Step 4: Apply the initial condition y'(π/2) = 2.
Substitute x = π/2 and y' = 2 into the first derivative of the general solution y'.
2 = -3cos(π/2) + A
2 = -3 + A (since cos(π/2) = 0)
Step 5: Solve the system of equations formed from Step 3 and Step 4 to find the values of A and B.
Using the second equation A = 5.
Substitute A = 5 into the first equation to find B.
π = -3 + (5π)/2 + B
π + 3 = (5π)/2 + B
B = π + 3 - (5π)/2
B = (2π + 6 - 5π)/2
B = (π + 6)/2
Step 6: Substitute the values of A and B into the general solution y = -3sin(x) + Ax + B to obtain the particular solution.
Therefore, the particular solution to the given differential equation y" = 3sin(x) with the initial conditions y(π/2) = π and y'(π/2) = 2 is:
y = -3sin(x) + 5x + (π + 6)/2
To find the particular solution to y"=3sin(x) given the general solution y=-3sin(x) +Ax+B and the initial conditions y (π/2)=π and y'(π/2)=2, we need to substitute the initial conditions into the general solution and solve for the unknown coefficients A and B.
First, let's substitute the initial condition y(π/2) = π into the general solution:
π = -3sin(π/2) + A(π/2) + B
Simplifying this equation:
π = -3(1) + A(π/2) + B
π = -3 + (Aπ)/2 + B
Next, let's substitute the initial condition y'(π/2) = 2 into the general solution:
2 = -3cos(π/2) + A
2 = -3(0) + A
2 = A
Now, we have two equations:
Equation 1: π = -3 + (Aπ)/2 + B
Equation 2: A = 2
Substituting A = 2 into Equation 1:
π = -3 + (2π)/2 + B
π = -3 + π + B
Simplifying this equation:
B = π + 3
Therefore, the particular solution to y"=3sin(x) given the general solution y=-3sin(x) +Ax+B and the initial conditions y (π/2)=π and y'(π/2)=2 is:
y = -3sin(x) + 2x + (π+3)