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Question

Show that the differential equation
x.cos(yx)dydx = y.cos(yx) + x
is homogeneous and solve it.





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Solution

Given: x.cos(yx)dydx = y.cos(yx) + x

dydx=ycos(yx)+xxcos(yx)

If F (λx,λy)=λF(x,y),
then the differential equation is homogeneous .

Substituting F(x,y)=ycos(yx)+xxcos(yx)
Now,

F(λx,λy)=(λy)cos(λyλx)+λx(λy).cos(λyλx)

F(λx,λy)=λycos(yx)+λxλycos(yx)

F(λx,λy=λ(y cos(yx)+x)λ x cos(yx)

F(λx,λy)= ycos(yx)+xxcos(yx)=F(x,y)


Hence,the differential equation is homogeneous.

Given differential equation is

x.cos(yx)dydx=y.cos(yx)+x

dydx=ycos(yx)+xxcos(yx) ...(i)

Lety=vx

Differentiating both sides w.r.t. 'x'.

dydx=d(vx)dx

Using product rule : (uv)=uv+uv

dydx=dvdxx+vdxdx

dydx=dvdxx+v

Substituting value of dydxand y=vx in (i),

dvdxx+v=(vx)cos(vxx)+xxcos(vxx)

dvdxx+v=x(vcos v+1)xcos v

dvdxx+v=v cos v+1x cos v

dvdxx=v cos v+1cos vv

dvdxx=v cos v+1v cos vcos v

dvdxx=1cos v

cos vdv=dxx

Integrating both sides, we get

cos v dv= dxx

sin v=log|x|+log|c|

Substituting v=yx,we get

sinyx=log|x|+log|c|

sinyx=log|cx| (log +log b=log ab)

The required general solution is

sinyx=log|cx|

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