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Solving a Riccati Differential Equation

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 5, 2012
1,621
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,

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