Question

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

Answer 1

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

$\displaystyle y^{1/2}\frac{\mathrm dy}{\mathrm dx} + y^{3/2} = 1$

Substitute

$z(x)=y(x)^{3/2} \implies \dfrac{\mathrm dz}{\mathrm dx}=\dfrac32y(x)^{1/2}\dfrac{\mathrm dy}{\mathrm dx}$

to transform the ODE to a linear one in z :

$\displaystyle \frac23\frac{\mathrm dz}{\mathrm dx} + z = 1$

Divide both sides by 2/3 :

$\displaystyle \frac{\mathrm dz}{\mathrm dx} + \frac32z = \frac32$

Multiply both sides by the integrating factor, $e^{3x/2}$ :

$\displaystyle e^{3x/2}\frac{\mathrm dz}{\mathrm dx} + \frac32 e^{3x/2}z = \frac32 e^{3x/2}$

Condense the left side into the derivative of a product :

$\displaystyle \frac{\mathrm d}{\mathrm dx}\left[e^{3x/2}z\right] = \frac32 e^{3x/2}$

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

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