How to solve $2c x + x\dot x + (c^2-1)t = cd$ After computation on a certain problem, for a specific case I came across the differential equation $$2c x + x \dot x + (c^2-1) t = cd$$
with initial conditions $0 < x(0) = d $ and  $1 < - \dot x(0) = c $. Does this have a closed form?
This differential equation doesn't appear to conform to solution by factors, integrating factors, Clairaut's equation, homogeneous methods, differentiating methods, Riccati's equation, or any other special form or method I know of. 
 A: $$2c x + \frac{dx}{dt} x + (c^2-1) t =cD$$
The symbol $D$ is used instead of $d$ to avoid confusion with the differential symbol.
Let $\quad cD=a$
$$2c x + \frac{dx}{dt} x + (c^2-1) t - a=0$$
$y(t)=(c^2-1)t-a  \quad;\quad  \frac{dx}{dt}= \frac{dx}{dy}\frac{dy}{dt}= (c^2-1)\frac{dx}{dy}$
$$2c x + (c^2-1)x\frac{dx}{dy} + y=0$$
$x=yu(y)\quad;\quad \frac{dx}{dy}=u+y\frac{du}{dy}$ 
$2c yu + (c^2-1)yu(u+y\frac{du}{dy}) + y=0$
$$2c u + (c^2-1)(u^2+yu\frac{du}{dy}) + 1=0$$
$y=e^z$
$$2c u + (c^2-1)(u^2+u\frac{du}{dz}) + 1=0$$
$$\frac{du}{dz} =-\frac{1+2cu}{(c^2-1)u}-u=-\frac{1+2cu+(c^2-1)u^2}{(c^2-1)u}=-\frac{ ((c+1)u+1)((c-1)u+1) }{(c^2-1)u}$$
$$z=(c^2-1)\int \frac{u}{((c+1)u+1)((c-1)u+1)}du+c_1$$
$$z=\frac{1}{2}(c+1)\ln|(c-1)u+1|-\frac{1}{2}(c-1)\ln|(c+1)u+1|+\text{constant}$$
$$y=e^z=c_1\frac{ ((c-1)u+1)^{(c+1)/2} }{((c+1)u+1)^{(c-1)/2}}$$
$$y=c_1\frac{ ((c-1)\frac{x}{y}+1)^{(c+1)/2} }{((c+1)\frac{x}{y}+1)^{(c-1)/2}}$$
$$y^2=c_1\frac{ ((c-1)x+y)^{(c+1)/2} }{((c+1)x+y)^{(c-1)/2}}$$
$$( (c^2-1)t-a) ^2=c_1\frac{ ((c-1)x+(c^2-1)t-a))^{(c+1)/2} }{((c+1)x+(c^2-1)t-a))^{(c-1)/2}}$$
The general solution of the ODE is expressed on the form of an implicit equation (with $a=cD$):
$$( (c^2-1)t-cD) ^2=c_1\frac{ ((c-1)x+(c^2-1)t-cD))^{(c+1)/2} }{((c+1)x+(c^2-1)t-cD))^{(c-1)/2}}$$
Condition $x(0)=D\quad\to\quad (-cd) ^2=c_1\frac{ ((c-1)D-cD))^{(c+1)/2} }{((c+1)D-cD))^{(c-1)/2}}\quad$ After simplification : $c_1=c^2D$
$$( (c^2-1)t-cD) ^2=c^2D\frac{ ((c-1)x+(c^2-1)t-cD))^{(c+1)/2} }{((c+1)x+(c^2-1)t-cD))^{(c-1)/2}}$$
This is the solution of the problem expressed on the form of implicit equation.
In case of first order ODE, only one boundary condition is sufficient. It is up to you to add an extra condition $\left(\frac{dx}{dt}\right)_{t=0}=-c$. Just test if it is compatible with the solution already found. If not, the problem would have no solution according to two boundary conditions instead of only one.
In the present case, at $t=0$ we have $x=D$ and $\frac{dx}{dt}=-c$. Putting them into the ODE gives : $2cD-cD=cD$ which is correct. So, the extra condition is implicitly satisfied. No need to worry about it.
