give the solution of the initial-value problem
dy/dx=(1+2cos^2(x))/y, (y > 0), y= 1 when x = 0.
Thanks.
Is it an answer on my question?
2cos^2(x)=1+cos(2x)
ydy=(2+cos(2x))dx integrating
y^2/2=2x+sin(2x)/2+C Find C
1/2=C
y=sqrt(4x+sin(2x)+1)
Thats right.
Thank you.
To solve the initial-value problem, we will first separate the variables and then integrate.
Given: dy/dx = (1 + 2cos^2(x))/y and y = 1 when x = 0.
Step 1: Separate the variables.
Rearrange the equation to have all y terms on one side and all x terms on the other side:
y * dy = (1 + 2cos^2(x)) * dx.
Step 2: Integrate both sides.
Integrate both sides of the equation separately:
∫ y * dy = ∫ (1 + 2cos^2(x)) * dx.
Integrating the left side gives:
(1/2)y^2 + C1,
where C1 is the constant of integration.
Integrating the right side requires the use of a trigonometric identity:
∫ (1 + 2cos^2(x)) * dx = x + sin(2x) + C2,
where C2 is another constant of integration.
Step 3: Combine the results to find the general solution.
Setting up the equation with both results:
(1/2)y^2 + C1 = x + sin(2x) + C2.
Step 4: Apply the initial condition.
Using the given initial condition where y = 1 when x = 0:
(1/2)1^2 + C1 = 0 + sin(2*0) + C2,
1/2 + C1 = 0 + 0 + C2,
C1 = C2 - 1/2.
Step 5: Write the final solution.
Substitute the value of C1 into the general solution:
(1/2)y^2 + (C2 - 1/2) = x + sin(2x) + C2.
This is the solution to the initial-value problem.