# Solving a Riccati Differential Equation

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Continuing from >>this<< thread, my friend gave me another question. He wants to check his work specifically on parts a) and d). So I will only focus on those parts.

Question Summary:

Given the Riccati equation,

$\frac{dy}{dx}+e^{-x}y^2+y+e^x=0$

we have to solve it using the substitution,

$y(x)=\frac{e^x}{w}\frac{dw}{dx}$

a) Show that,

$\frac{dy}{dx}=e^x\left[\frac{1}{w}\frac{d^{2}w}{dx^2}-\frac{1}{w^2}\left(\frac{dw}{dx}\right)^2+\frac{1}{w}\frac{dx}{dx}\right]$

b) Show the differential equation for $$w(x)$$ is,

$\frac{d^{2}w}{dx^2}+2\frac{dw}{dx}+w=0$

c) Find the general solution of the differential equation you found in part b).

d) Use $$\displaystyle y(x)=\frac{e^x}{w}\frac{dw}{dx}$$ to find the general solution of $$y(x)$$ of the original Riccati equation. Simplify this expression as much as possible. You should be able to combine the two integration constants from part c) into one single integration constant.

#### Sudharaka

##### Well-known member
MHB Math Helper
For part a),

Use the product rule of differentiation.

$y(x)=\frac{e^x}{w}\frac{dw}{dx}$
\begin{eqnarray}

\frac{dy}{dx}&=&\frac{e^x}{w}\frac{d^{2}w}{dx^2}+ \frac{dw}{dx}\frac{d}{dx}\left(\frac{e^x}{w}\right)\\

&=&\frac{e^x}{w}\frac{d^{2}w}{dx^2}+ \frac{dw}{dx} \left(\frac{e^x}{w}-\frac{e^x}{w^2}\frac{dw}{dx}\right)\\

&=&e^x\left(\frac{1}{w}\frac{d^{2}w}{dx^2}-\frac{1}{w^2}\left(\frac{dw}{dx}\right)^2+\frac{1}{w}\frac{dw}{dx} \right)\\

\end{eqnarray}

For part b),

Substitute $$\displaystyle y(x)=\frac{e^x}{w}\frac{dw}{dx}$$ and the above result we obtained for $$y'(x)$$ in the Riccati equation. Simplify the get the desired result.

For part c),

This is a second order, linear, homogeneous differential equation with constant coefficients. Therefore you can use the trial solution $$w(x)=Ae^{mx}$$ where $$m$$ is to be determined. The characteristic equation will be,

$m^2+2m+1=0\Rightarrow m=-1$

Therefore,

$w(x)=(A+Bx)e^{-x}$

where $$A$$ and $$B$$ are arbitrary constants.

For part d),

Substituting the above result in $$\displaystyle y(x)=\frac{e^x}{w}\frac{dw}{dx}$$ we get,

$y(x)=\frac{e^x}{(A+Bx)e^{-x}}\frac{d}{dx}\left[(A+Bx)e^{-x}\right]$

$\Rightarrow y(x)=\left(\frac{B-A-Bx}{A+Bx}\right)e^{x}$

Dividing the denominator and numerator by $$B$$ and substituting $$\displaystyle C=\frac{A}{B}$$ we get,

$y(x)=\left(\frac{1-C-x}{C+x}\right)e^{x}=\left(\frac{1}{C+x}-1\right)e^{x}$