y′=12x^(3)sqrt(1+x^4)(1+y^2), y(0)=1
⇒ y=
Thanks!
y′=12x^3 √(1+x^4)(1+y^2)
Man, after the last one, this should not have caused you any headaches. It's basically the same thing
dy = 12x^3 √(1+x^4)(1+y^2) dx
dy/(1+y^2) = 12x^3 √(1+x^4)
arctan(y) = 2(1+x^2)^(3/2) + c
y = tan(2(1+x^2)^(3/2) + c)
y(0)=1 ⇒ tan(2+c) = 1 ⇒ 2+c = π/4
y = tan(2(1+x^2)3/2 + π/4-2)
Better check my math
oops typo that should be √(1+x^4) everywhere. I seem to have slipped in some 1+x^2...
To find the solution to this differential equation, we'll use the method of separation of variables.
Step 1: Separate the variables
Start by writing the equation in the form:
dy / dx = 12x^3 * sqrt(1 + x^4) * (1 + y^2)
Step 2: Move the terms involving y to one side and the terms involving x to the other side.
Divide both sides by (1 + y^2):
(1 + y^2) dy = 12x^3 * sqrt(1 + x^4) dx
Step 3: Integrate both sides
Integrate both sides of the equation:
∫ (1 + y^2) dy = ∫ 12x^3 * sqrt(1 + x^4) dx
For the left side, we can use the formula for integrating powers of y:
∫ (1 + y^2) dy = y + (y^3 / 3)
For the right side, we can use a substitution to simplify the integral. Let u = 1 + x^4.
Then, du = 4x^3 dx, which we can rearrange to get dx = (1/4x^3) du.
Substituting back into the equation, we get:
∫ 12x^3 * sqrt(1 + x^4) dx = 12 * ∫ (1/4x^3) * sqrt(u) du = 3 * ∫ u^(-1/2) du = 3 * 2sqrt(u) = 6sqrt(u)
So our equation becomes:
y + (y^3 / 3) = 6sqrt(u)
Step 4: Replace u with 1 + x^4
Substituting back, we have:
y + (y^3 / 3) = 6sqrt(1 + x^4)
Step 5: Solve for y
Now we need to solve this equation for y. Since the initial condition is given as y(0) = 1, we can substitute x = 0 and y = 1 into the equation:
1 + ((1)/3) = 6sqrt(1 + (0)^4)
1 + (1/3) = 6sqrt(1)
Simplifying, we get:
4/3 = 6
This is a contradiction, so there is no solution to this initial value problem.