Answer to Question #107631 in Differential Equations for Nikesh gautam pandit ji

Question

Find the value of b for which the equation
(ye^2xy + x) dx + (bxe^2xy) dy =0
is exact, and hence solve it for that value of b

b) Find the solution of the Riccati equation
dy/dx =2cos^2x-sin^2x+y^2 / 2cosx
y1(x) = sin x

Answer 1

a) $(ye^{2xy} + x) dx + (bxe^{2xy}) dy=dU$, then $ye^{2xy}+x=\frac{\partial U}{\partial x}$ and $bxe^{2xy}=\frac{\partial U}{\partial y}$.

We have $\frac{\partial^2 U}{\partial x\partial y}=\frac{\partial^2 U}{\partial y\partial x}$, that is $\frac{\partial}{\partial x}(bxe^{2xy})=\frac{\partial}{\partial y}(ye^{2xy}+x)$

$\frac{\partial}{\partial x}(bxe^{2xy})=be^{2xy}+2bxye^{2xy}$, $\frac{\partial}{\partial y}(ye^{2xy}+x)=e^{2xy}+2xye^{2xy}$ , so $be^{2xy}+2bxye^{2xy}=e^{2xy}+2xye^{2xy}$, and we obtain $b=1$.

Hence $\frac{\partial U}{\partial y}=xe^{2xy}$, so $U=\frac{e^{2xy}}{2}+f(x)$.

Then $\frac{\partial U}{\partial x}=ye^{2xy}+f'(x)$. Since $\frac{\partial U}{\partial x}=ye^{2xy}+x$, we have $f'(x)=x$.

Let $f(x)=\frac{x^2}{2}$, then $U=\frac{e^{2xy}+x^2}{2}$

So the given equation we can rewrite as $d\left(\frac{e^{2xy}+x^2}{2}\right)=0$, general solution of which is $\frac{e^{2xy}+x^2}{2}=\frac{C}{2}$ or $e^{2xy}+x^2=C$

b)General solution we will look in form $y=y_1+u=u+\sin x$

We have $y'=u'+\cos x$, then

$u'+\cos x=\frac{2\cos^2x-\sin^2x+(u+\sin x)^2}{2cosx}$

After expanding brackets and moving $\cos x$ to the right side, we obtain

$u'=\frac{u^2+2u\sin x}{2cosx}$ or $\frac{du}{dx}=\frac{u^2+2u\sin x}{2cosx}$

So $2\cos xdu=u^2dx+2u\sin xdx$ or $2\cos xdu=u^2dx-2ud(\cos x)$

Since $d(UV)=VdU+UdV$ for every differentiable $U$ and $V$, we have $2d(u\cos x)=u^2dx$

Let $u\cos x=v$, then $2dv=\frac{v^2}{\cos^2 x}dx$.

So we have $v=0$ or $\frac{2dv}{v^2}=\frac{dx}{\cos^2 x}$ or $-2d\left(\frac{1}{v}\right)=d(\tan x)$ . We obtain the general solution $v=0$ or $\tan x+\frac{2}{v}=C$.

Reurning to $u$, we obtain $u=0$ or $\tan x+\frac{2}{u\cos x}=C$. And, returning to $y$, we obtain $y=\sin x$ or $\tan x+\frac{2}{(y-\sin x)\cos x}=C$. Expessing $y$, we obtain $y=\sin x$ or $y=\sin x+\frac{2}{C\cos x-\sin x}$

Answer: a)$b=1$, $e^{2xy}+x^2=C$, b)$y=\sin x$ or $y=\sin x+\frac{2}{C\cos x-\sin x}$