Choose the option that gives the solution of the initial-value problem
dy
dx=(1 + 2 cos2x^2)/y (y > 0), y= 1 when x = 0.
Options
A y = 1+ 2sin^2 x
B y = (1 + 2 sin x)^2
C y = (4x + cos(2x))^2
D y = 4x + cos^2 x
E y =sqrt(4x + cos(2x)
F y =sqrt(1 + 4x + sin(2x))
G y =1/(sqrt(2x + cos(2x))
H y = 1/(sqrt(1 + sin(2x))
To find the solution of the initial-value problem, we need to solve the given differential equation and then apply the initial condition.
The given differential equation is:
dy/dx = (1 + 2cos^2(x^2))/y
To solve this, we can start by separating the variables. Rearrange the equation to have y terms on one side and x terms on the other side:
y dy = (1 + 2cos^2(x^2)) dx
Next, integrate both sides with respect to their respective variables:
∫ y dy = ∫ (1 + 2cos^2(x^2)) dx
Integrating the left side:
(1/2) y^2 = x + 2∫ cos^2(x^2) dx
To tackle the integral on the right side, we can use the trigonometric identity:
cos^2(x^2) = (1 + cos(2x^2))/2
Substituting this into the integral:
(1/2) y^2 = x + 2∫ (1 + cos(2x^2))/2 dx
(1/2) y^2 = x + ∫ (1/2) dx + ∫ cos(2x^2)/2 dx
(1/2) y^2 = x + (1/2)x + (1/4)∫ cos(2x^2) dx
To solve the integral of cos(2x^2), we need to use a change of variables. Let u = 2x^2, then du = 4x dx. Rearranging and substituting back into the equation, we get:
(1/2) y^2 = x + (1/2)x + (1/4)∫ cos(u)/4 du
(1/2) y^2 = x + (1/2)x + (1/16)∫ cos(u) du
Integrating cos(u) with respect to u, we get:
(1/2) y^2 = x + (1/2)x + (1/16)sin(u) + C
Substituting back u = 2x^2:
(1/2) y^2 = x + (1/2)x + (1/16)sin(2x^2) + C
Now, apply the initial condition y = 1 when x = 0:
(1/2) (1)^2 = 0 + (1/2)(0) + (1/16)sin(2(0)^2) + C
(1/2) = 0 + 0 + 0 + C
C = 1/2
Therefore, the solution to the initial-value problem is:
(1/2) y^2 = x + (1/2)x + (1/16)sin(2x^2) + (1/2)
Simplifying:
y^2 = 2x + x + (1/8)sin(2x^2) + 1
y^2 = 3x + (1/8)sin(2x^2) + 1
Taking the square root of both sides (since y > 0):
y = sqrt(3x + (1/8)sin(2x^2) + 1)
Therefore, the answer is option E: y = sqrt(3x + (1/8)sin(2x^2) + 1).