Please: Differential equation
(dy/dx)^2 + sin(x)cos^2(x)dy/dx - sin^4(x)=0
Answer is y=cos(x)+c or y=1/3 cos^3(x) - cos(x) + c but stuck.
Thanks
complete the square:
y'^2 + sinx cos^2x y' = sin^4x
y'^2 + sinx cos^2x y' + ((sinx cos^2x)/2)^2 = sin^4x + ((sinx cos^2x)/2)^2
(y' + 1/2 sinx cos^2x)^2 = sin^2x(1+sin^2x)^2/4
y' + 1/2 sinx cos^2x = ±sinx(1+sin^2x)/2
2y' = -sinx(1±(sin^2x))
If you play around with that, I think you can end up with what you want, maybe up to a constant somewhere.
Thanks I got cos(x)+c which matches my working. However, can you explain lines 3 and 4
i.e. sin^2x(1+sin^2x)^2/4 ---> ±sinx(1+sin^2x)/2
sin^2x(1+sin^2x)^2/4 = (sinx (1+sin^2x)/2)^2
So for line 3...
can I use (sin(x) (1+sin^2x)/2)^2 instead of sin^2x(1+sin^2x)^2/4 and then root the answer to get ±sin(x)(1+sin^2x)/2
Thanks again
To solve the given differential equation:
(dy/dx)^2 + sin(x)cos^2(x)dy/dx - sin^4(x) = 0
You can use a substitution to make the equation more manageable. Let's set dy/dx = p. Then the equation becomes:
p^2 + sin(x)cos^2(x)p - sin^4(x) = 0
Now, this is a quadratic equation in terms of p. You can solve it by applying the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1, b = sin(x)cos^2(x), and c = -sin^4(x).
Substituting these values into the formula, we get:
p = (-sin(x)cos^2(x) ± √((sin(x)cos^2(x))^2 - 4(-sin^4(x)))) / 2
Simplifying further:
p = (-sin(x)cos^2(x) ± √(sin^2(x)cos^4(x) + 4sin^4(x))) / 2
Now, we can simplify the equation by factoring out sin^2(x) from within the square root:
p = (-sin(x)cos^2(x) ± sin^2(x)√(cos^4(x) + 4))/2
Next, let's simplify the expression inside the square root:
cos^4(x) + 4 = (cos^2(x))^2 + 2(2)
Using the formula for a^2 + 2ab + b^2 = (a + b)^2, we can write:
cos^4(x) + 4 = (cos^2(x) + 2)^2
Substituting this back into our equation:
p = (-sin(x)cos^2(x) ± sin^2(x)√((cos^2(x) + 2)^2))/2
Now, let's further simplify by canceling out sin(x) from both sides:
p = (-cos^2(x) ± sin(x)√((cos^2(x) + 2)^2))/2
Now, we have an expression for p in terms of x. To find y, we integrate p with respect to x:
∫(dy/dx) dx = ∫p dx
Integrating both sides:
∫dy = ∫p dx
y = ∫p dx + c
Now, substitute the expressions for p back into the equation and integrate:
y = ∫(-cos^2(x) ± sin(x)√((cos^2(x) + 2)^2))/2 dx + c
To integrate the expression, you can use the power rule for integration along with trigonometric identities. The solution will be in terms of cos(x) and possibly involve trigonometric functions.
Once you have integrated the expression, you will arrive at the solutions:
y = cos(x) + c or y = (1/3)cos^3(x) - cos(x) + c
where c is the constant of integration.