2.1.11.5 Solved using first_order_ode_dAlembert

Entering first order ode dAlembert solver

\[\begin {aligned} x^{\prime }&=2 \sqrt {x}\\ x \left (0\right ) &= 1\\ \end {aligned}\]

Let \(p=x^{\prime }\) the ode becomes

\begin{align*} p = 2 \sqrt {x} \end{align*}

Solving for \(x\) from the above results in

\begin{align*} x &= \frac {p^{2}}{4}\tag {1A} \end{align*}

This has the form

\begin{align*} x&=t f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=x'(t)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(t\) gives

\begin{align*} p &= f+(t f'+g') \frac {dp}{dt}\\ p-f &= (t f'+g') \frac {dp}{dt}\tag {2} \end{align*}

Comparing the form \(x=t f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= \frac {p^{2}}{4} \end{align*}

Hence (2) becomes

\begin{align*} p = \frac {p p^{\prime }\left (t \right )}{2}\tag {3} \end{align*}

The singular solution is found by setting \(\frac {dp}{dt}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}t}}\neq 0\) from eq. (3). This results in

\begin{align*} p^{\prime }\left (t \right ) = 2\tag {4} \end{align*}

This ODE is now solved for \(p \left (t \right )\). No inversion is needed.

Because the ODE has the form \(p^{\prime }\left (t \right )=f(t)\), the solution requires only integration. Therefore

\begin{align*} dp &= \left (2\right ) \, dt\\ p &= \int { \left (2\right ) \, dt}\\ &= 2 t +c_1 \end{align*}

Substituing the above solution for \(p\) in \(x = \frac {p^{2}}{4}\) gives

\[ x = \frac {\left (2 t +c_1 \right )^{2}}{4} \]

Initial condition \(x \left (0\right ) = 1\) is now applied. Applying the initial condition \(x \left (0\right ) = 1\), the solution becomes

\begin{align*} x &= \frac {\left (2 t -2\right )^{2}}{4} \\ x &= \frac {\left (2 t +2\right )^{2}}{4} \\ \end{align*}

Simplifying \(x = \frac {\left (2 t -2\right )^{2}}{4}\) gives

\[ x = \left (-1+t \right )^{2} \]

Simplifying \(x = \frac {\left (2 t +2\right )^{2}}{4}\) gives

\[ x = \left (t +1\right )^{2} \]

Direction field with Isoclines	Direction field and Solutions plot

Summary of solutions found

\begin{align*} x&=\left (-1+t \right )^{2}\\ x&=\left (t +1\right )^{2} \end{align*}