# Thread: Differential equations know one solution

1. ## Differential equations know one solution

1) Solve :

$y' = \cos (x) - y\sin (x) + y^2
$

Know : $y = \sin (x)$ is the solution.

2 ) Solve :

$y' = 2 + \frac{1}{2}\left( {x - \frac{1}{x}} \right)y - \frac{1}{2}y^2
$

Know :
$y = x + \frac{1}{x}$ is the solution.

2. Ricatti's equations. There's no much here. What have you done?

3. Originally Posted by dhiab
1) Solve :

$y' = \cos (x) - y\sin (x) + y^2
$

Know : $y = \sin (x)$ is the solution.
If we know a particular solution $y_p$ to the Riccati equation:

$y'+Q(x)y+R(x)y^2=P(x)$

then we can obtain the general solution by making the substitution:

$y=\frac{1}{u}+y_p$

and obtain an easier first order equation:

$u'-(2Ry_p+Q)u=R$

Now, when I do that for the first one, I obtain the general solution for this equation:

$y(x)=\sin(x)+\left(c_1 e^{\cos(x)}-e^{\cos(x)}\int e^{-\cos(x)}\right)^{-1}$.

Now, the interesting part I think, is back-substituting that solution into the original Riccati and verifying that it satisfies it. Forgive me if I'm a little lazy and just used Mathematica:

Code:
In[61]:=
y[x_] := Sin[x] + 1/(c1*Exp[Cos[x]] -
Exp[Cos[x]]*Integrate[Exp[-Cos[x]],
x]);
FullSimplify[D[y[x], x] + Sin[x]*y[x] -
y[x]^2]

Out[62]= Cos[x]
I'm sure you could do the second one now.