Question

"Solve the following first order differential equation for x(t):

x'=-9tx"


How do I do this?

Answer 1

I'm not sure if the last two apostrophes are part of the quote - "Solve ... " - or if you mean the second derivative
`x''`. I think you mean the first interpretation, but I'll include both cases since they are both solvable.

If the former is correct, separate variables to solve.

`x' = -9tx \implies (dx)/(dt) = -9tx \implies \frac{dx}x = -9t\,dt`

Integrate both sides to get

`\ln|x| = -\frac92 t^2 + C`

Solve for
`x`.

`e^(\ln|x|) = e^(-9/2\,t^2 + C) \implies \boxed{x = Ce^(-9/2\,t^2)}`

If you meant the latter, then the ODE can be rewritten as

`9t x'' + x' = 0`

Reduce the order of the equation by substituting
`y(t) = x'(t)` and
`y'(t) = x''(t)`.

`9t y' + y = 0`

Solve for
`y'` and separate variables.

`y' = -\frac y{9t} \implies (dy)/(dt) = -\frac y{9t} \implies \frac{dy}y = -(dt)/(9t)`

Integrate.

`\ln|y| = -\frac19 \ln|t| + C`

Solve for
`y`.

`e^(\ln|y|) = e^(-1/9 \,\ln|t| + C) \implies y = Ct^(-1/9)`

Solve for
`x` by integrating.

`x' = Ct^(-1/9) \implies x = C_1 t^(8/9) + C_2`