Question

Solve the given initial-value problem. The DE is a Bernoulli equation.

, 1/2 dy + y3/2 = 1, y(0) = 16
dx
2
3
x
2
|
co
y = 1 + 63e
x
Show all work correctly

Answer 1

Your solution seems fine. What does the rest of the error message say?

`\displaystyle y^(1/2)(\mathrm dy)/(\mathrm dx) + y^(3/2) = 1`

Substitute

`z(x)=y(x)^(3/2) \implies (\mathrm dz)/(\mathrm dx)=\frac32y(x)^(1/2)(\mathrm dy)/(\mathrm dx)`

to transform the ODE to a linear one in z :

`\displaystyle \frac23(\mathrm dz)/(\mathrm dx) + z = 1`

Divide both sides by 2/3 :

`\displaystyle (\mathrm dz)/(\mathrm dx) + \frac32z = \frac32`

Multiply both sides by the integrating factor,
`e^(3x/2)` :

`\displaystyle e^(3x/2)(\mathrm dz)/(\mathrm dx) + \frac32 e^(3x/2)z = \frac32 e^(3x/2)`

Condense the left side into the derivative of a product :

`\displaystyle (\mathrm d)/(\mathrm dx)\left[e^(3x/2)z\right] = \frac32 e^(3x/2)`

Integrate both sides and solve for z :

`\displaystyle e^(3x/2)z = \frac32 \int e^(3x/2)\,\mathrm dx \\\\ e^(3x/2)z = e^(3x/2) + C \\\\ z = 1 + Ce^(-3x/2)`

Solve in terms of y :

`y^(3/2) = 1 + Ce^(-3x/2)`

Given that y (0) = 16, we have

`16^(3/2) = 1 + Ce^0 \implies C = 16^(3/2)-1 = 63`

so that the particular solution is

`\boxed{y^(3/2) = 1 + 63e^(-3x/2)}`