Question #107631
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
1
Expert's answer
2020-04-02T13:46:55-0400

a) (ye2xy+x)dx+(bxe2xy)dy=dU(ye^{2xy} + x) dx + (bxe^{2xy}) dy=dU, then ye2xy+x=Uxye^{2xy}+x=\frac{\partial U}{\partial x} and bxe2xy=Uybxe^{2xy}=\frac{\partial U}{\partial y}.

We have 2Uxy=2Uyx\frac{\partial^2 U}{\partial x\partial y}=\frac{\partial^2 U}{\partial y\partial x}, that is x(bxe2xy)=y(ye2xy+x)\frac{\partial}{\partial x}(bxe^{2xy})=\frac{\partial}{\partial y}(ye^{2xy}+x)

x(bxe2xy)=be2xy+2bxye2xy\frac{\partial}{\partial x}(bxe^{2xy})=be^{2xy}+2bxye^{2xy}, y(ye2xy+x)=e2xy+2xye2xy\frac{\partial}{\partial y}(ye^{2xy}+x)=e^{2xy}+2xye^{2xy} , so be2xy+2bxye2xy=e2xy+2xye2xybe^{2xy}+2bxye^{2xy}=e^{2xy}+2xye^{2xy}, and we obtain b=1b=1.

Hence Uy=xe2xy\frac{\partial U}{\partial y}=xe^{2xy}, so U=e2xy2+f(x)U=\frac{e^{2xy}}{2}+f(x).

Then Ux=ye2xy+f(x)\frac{\partial U}{\partial x}=ye^{2xy}+f'(x). Since Ux=ye2xy+x\frac{\partial U}{\partial x}=ye^{2xy}+x, we have f(x)=xf'(x)=x.

Let f(x)=x22f(x)=\frac{x^2}{2}, then U=e2xy+x22U=\frac{e^{2xy}+x^2}{2}

So the given equation we can rewrite as d(e2xy+x22)=0d\left(\frac{e^{2xy}+x^2}{2}\right)=0, general solution of which is e2xy+x22=C2\frac{e^{2xy}+x^2}{2}=\frac{C}{2} or e2xy+x2=Ce^{2xy}+x^2=C

b)General solution we will look in form y=y1+u=u+sinxy=y_1+u=u+\sin x

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

u+cosx=2cos2xsin2x+(u+sinx)22cosxu'+\cos x=\frac{2\cos^2x-\sin^2x+(u+\sin x)^2}{2cosx}

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

u=u2+2usinx2cosxu'=\frac{u^2+2u\sin x}{2cosx} or dudx=u2+2usinx2cosx\frac{du}{dx}=\frac{u^2+2u\sin x}{2cosx}

So 2cosxdu=u2dx+2usinxdx2\cos xdu=u^2dx+2u\sin xdx or 2cosxdu=u2dx2ud(cosx)2\cos xdu=u^2dx-2ud(\cos x)

Since d(UV)=VdU+UdVd(UV)=VdU+UdV for every differentiable UU and VV, we have 2d(ucosx)=u2dx2d(u\cos x)=u^2dx

Let ucosx=vu\cos x=v, then 2dv=v2cos2xdx2dv=\frac{v^2}{\cos^2 x}dx.


So we have v=0v=0 or 2dvv2=dxcos2x\frac{2dv}{v^2}=\frac{dx}{\cos^2 x} or 2d(1v)=d(tanx)-2d\left(\frac{1}{v}\right)=d(\tan x) . We obtain the general solution v=0v=0 or tanx+2v=C\tan x+\frac{2}{v}=C.

Reurning to uu, we obtain u=0u=0 or tanx+2ucosx=C\tan x+\frac{2}{u\cos x}=C. And, returning to yy, we obtain y=sinxy=\sin x or tanx+2(ysinx)cosx=C\tan x+\frac{2}{(y-\sin x)\cos x}=C. Expessing yy, we obtain y=sinxy=\sin x or y=sinx+2Ccosxsinxy=\sin x+\frac{2}{C\cos x-\sin x}


Answer: a)b=1b=1, e2xy+x2=Ce^{2xy}+x^2=C, b)y=sinxy=\sin x or y=sinx+2Ccosxsinxy=\sin x+\frac{2}{C\cos x-\sin x}


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