Can any one help with the following please:
Solve the initial-value problem:
problem:
dy/dx: 4(3cos(6x)-1)/1+2x-sin(6x) y=12 when x=0
Do you mean
dy/dx= 4(3cos(6x)-1) / (1+2x-sin(6x)) ?
Substitute
u=(1+2x-sin(6x))
du=(-6cos(6x)+2)
so
dy/dx = -2∫du/u
y=-2ln(1+2x-sin(6x))+C
Initial condition: y(0)=12
y(0)=12=-2ln(1+2(0)-sin(6(0)))+C
12=-2ln(1)+C
12=0+C
=> C=12
So
y=-2ln(1+2x-sin(6x))+12
Check the result by differentiation.
To solve the initial-value problem, follow these steps:
Step 1: Rewrite the differential equation
dy/dx = [4(3cos(6x) - 1)] / [1 + 2x - sin(6x)]
Step 2: Separate variables
Multiply both sides of the equation by dx to get the y terms on one side and the x terms on the other side:
1 + 2x - sin(6x) dy = 4(3cos(6x) - 1) dx
Step 3: Integrate both sides
Integrate both sides of the equation with respect to x:
∫ (1 + 2x - sin(6x)) dy = ∫ 4(3cos(6x) - 1) dx
Step 4: Evaluate the integrals
Evaluate the integrals on both sides. The integral on the left side will be ∫ dy, which is simply y:
y = ∫ 4(3cos(6x) - 1) dx = 4∫ (3cos(6x) - 1) dx
To solve this integral, split it into two separate integrals:
y = 12∫ cos(6x) dx - 4∫ dx
Integrating each term gives:
y = 12 * (1/6)sin(6x) - 4x + C
where C is the constant of integration.
Step 5: Substitute the initial condition
Since the problem provides the initial condition y = 12 when x = 0, substitute these values into the solution:
12 = 12 * (1/6)sin(6(0)) - 4(0) + C
12 = 0 + 0 + C
C = 12
Step 6: Finalize the solution
Now that you have the constant of integration, the final solution to the initial-value problem is:
y = 12 * (1/6)sin(6x) - 4x + 12