A differential equation question
Find the general solution to the equation
(dy/dx)^2 + sinxcos^2x(dy/dx) - sin^4x = 0
Thanks
(dy/dx)^2 + sinx cosx (dy/dx) - sin^4x = 0
You sure about that? Because --
http://www.wolframalpha.com/input/?i=(dy%2Fdx)%5E2+%2B+sinx+cosx+(dy%2Fdx)+-+sin%5E4x+%3D+0
On the other hand, if you meant
y" + sinx cosx y' = sin^4x
Then you can substitute u=y' and get
u' + sinx cosx u = sin^4x
Then use the integrating factor e^(∫sinx cosx dx) = e^(1/2 sin^2x) to solve for u, and then let y = ∫u du
I took a 2nd look at this, and it occurred to me how to solve what you had typed.
y'^2 + sinx cosx y' = sin^4x
y'^2 + sinx cosx y' + sin^2x cos^2x/4 = sin^4x + sin^2x cos^2x/4
4y'^2 + 4sinx cosx y' + sin^2x cos^2x = 4sin^4x + sin^2x cos^2x
(2y' + sinx cosx)^2 = sin^2x(4sin^2x-cos^2x)
2y' + sinx cosx = ±sinx √(1-3cos^2x)
4y' = -sin2x ±u√(1-3u^2) where u=cosx
y' = 1/2 cos2x ± 1/9 (1-3cos^2x)^(3/2) + c
Double-check. sometimes wolframalpha.com comes up with complicated and convoluted solutions.
To find the general solution to the given differential equation, we can follow these steps:
Step 1: Let's first consider the equation in terms of the derivative dy/dx: (dy/dx)^2 + sin(x)cos^2(x)(dy/dx) - sin^4(x) = 0.
Step 2: Now, observe that this equation is of the quadratic form. Let's substitute dy/dx with a new variable, say v, to simplify the equation. So, we have v^2 + sin(x)cos^2(x)v - sin^4(x) = 0.
Step 3: The above quadratic equation can be factored. Let's find the factors of sin(x)cos^2(x) and sin^4(x) so that we can rewrite the equation in factored form.
sin(x)cos^2(x)v^2 + sin^5(x) - sin^4(x)v = 0
sin^4(x)(cos^2(x)v^2 + sin(x) - v) = 0
Now, the equation becomes sin^4(x)(cos^2(x)v^2 + sin(x) - v) = 0.
Step 4: Since sin^4(x) is never equal to zero, we can divide both sides of the equation by sin^4(x) to get:
cos^2(x)v^2 + sin(x) - v = 0
Step 5: Now, we have a quadratic equation with respect to v. We can solve this quadratic equation by using the quadratic formula:
v = (-B ± sqrt(B^2 - 4AC)) / 2A
In our case, A = 1, B = sin(x), and C = -1. Plugging these values into the quadratic formula, we get:
v = (-sin(x) ± sqrt(sin^2(x) + 4cos^2(x))) / (2cos^2(x))
Step 6: Simplifying the expression inside the square root, we have:
sqrt(sin^2(x) + 4cos^2(x)) = sqrt(sin^2(x) + (2cos(x))^2)
= sqrt(sin^2(x) + (2cos^2(x) - 4sin^2(x)))
= sqrt(2cos^2(x) - 3sin^2(x))
Step 7: Substitute this expression back into the equation for v:
v = (-sin(x) ± sqrt(2cos^2(x) - 3sin^2(x))) / (2cos^2(x))
Hence, we obtain the general solution to the given differential equation:
dy/dx = (-sin(x) ± sqrt(2cos^2(x) - 3sin^2(x))) / (2cos^2(x))
This can also be expressed as:
dy/dx = (-sin(x) ± sqrt(2 - 3tan^2(x))) / (2sec^2(x))