2.1.2.3 Solved using first_order_ode_dAlembert

Entering first order ode dAlembert solver

\[\begin {aligned} y^{\prime }+c y&=a \end {aligned}\]

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

\begin{align*} c y +p = a \end{align*}

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

\begin{align*} y &= \frac {a -p}{c}\tag {1A} \end{align*}

This has the form

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

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

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

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

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

\begin{align*} f &= 0\\ g &= \frac {a -p}{c} \end{align*}

Hence (2) becomes

\begin{align*} p = -\frac {p^{\prime }\left (x \right )}{c}\tag {3} \end{align*}

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

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y&=\frac {a}{c} \end{align*}

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

\begin{align*} p^{\prime }\left (x \right ) = -p \left (x \right ) c\tag {4} \end{align*}

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

Integrating gives

\begin{align*} \int -\frac {1}{p c}d p &= dx\\ -\frac {\ln \left (p \right )}{c}&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} -p c&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right )&=0 \end{align*}

Solving for \(p\) from above gives

\[ p = \frac {-{\mathrm e}^{-c_1 c -x c}+a}{c} \]

Substituing the above solution for \(p\) in \(y = \frac {a -p}{c}\) gives

\[ y = \frac {-{\mathrm e}^{-c_1 c -x c}+a}{c} \]

Substituing the above solution for \(p\) in \(y = \frac {a -p}{c}\) gives

\[ y = \frac {a}{c} \]

Simplifying \(y = \frac {-{\mathrm e}^{-c_1 c -x c}+a}{c}\) gives

\[ y = \frac {-{\mathrm e}^{-c \left (x +c_1 \right )}+a}{c} \]

Summary of solutions found

\begin{align*} y&=\frac {a}{c}\\ y&=\frac {-{\mathrm e}^{-c \left (x +c_1 \right )}+a}{c} \end{align*}