- #1
1d20
- 12
- 0
I've been unable to fully solve this: [itex]\frac{dy}{dx} + y = xy^4[/itex]
The U: [itex]u = y^{-3}[/itex], so [itex]y = u^\frac{-1}{3}[/itex], and [itex]\frac{dy}{dx} = \frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} [/itex]
The substitution: [itex]\frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} + u^\frac{-1}{3} = xu^\frac{-4}{3}[/itex]
Simplified: [itex]\frac{du}{dx} - 3u = -3x[/itex]
AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex]
I've tried solving the resulting equation as a linear:
[itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex].
Which creates this unworkable equation: [itex]e^{\frac{3}{2}x^2}\frac{du}{dx} + 3xe^{\frac{3}{2}x^2} = 3ue^{\frac{3}{2}x^2}[/itex]
And I've tried solving it as a homogenous ([itex]3udx - 3xdx - du = 0[/itex]):
[itex]u = vx[/itex], so [itex]du = vdx + xdv[/itex]
Subing those in creates this unworkable mess: [itex]3vxdx - 3xdx - vdx - xdv = 0[/itex]
Which leaves me stuck, because I've only learned to solve separable, exact, homogenous, linear and bernoullis.
The U: [itex]u = y^{-3}[/itex], so [itex]y = u^\frac{-1}{3}[/itex], and [itex]\frac{dy}{dx} = \frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} [/itex]
The substitution: [itex]\frac{-1}{3}u^\frac{-4}{3}\frac{du}{dx} + u^\frac{-1}{3} = xu^\frac{-4}{3}[/itex]
Simplified: [itex]\frac{du}{dx} - 3u = -3x[/itex]
AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex]
I've tried solving the resulting equation as a linear:
[itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex].
Which creates this unworkable equation: [itex]e^{\frac{3}{2}x^2}\frac{du}{dx} + 3xe^{\frac{3}{2}x^2} = 3ue^{\frac{3}{2}x^2}[/itex]
And I've tried solving it as a homogenous ([itex]3udx - 3xdx - du = 0[/itex]):
[itex]u = vx[/itex], so [itex]du = vdx + xdv[/itex]
Subing those in creates this unworkable mess: [itex]3vxdx - 3xdx - vdx - xdv = 0[/itex]
Which leaves me stuck, because I've only learned to solve separable, exact, homogenous, linear and bernoullis.