Question

(y cos x+ 1) dx+sinxdy = 0​

Answer 1

Answer:

For y#0 the equation can be written as M(x,y)dx + N(x,y)dy =0 , with

M = ycosx , M_y = cosx

N = (y + 2)sinx , N_x = (y+2)cosx .The equation is not exact but

(N_x - M_y)/M = (y+1)/y depends only on y .The integrating factor ye^y

produces the equation P(x,y)dx + N(x,y)dy =0 with

P = y^2e^ycosx , P_y = (2ye^y + y^2e^y)cosx

Q = (y^2e^y + 2ye^y)sinx , Q_x = P_y. The equation is exact and is the total

differential dF(x,y) =0 whose solution F(x,y) =C is obtained by

F_x = P = y^2e^ycosx

F_y = Q = (y^2e^y + 2ye^y)sinx. Integrating the first Eq. leads to

F(x,y) = y^2e^ysinx + G(y). The function G(y) is obtained from the second Eq.

which gives G’(y) =0 .Then G(y)= C and the solution is

F(x,y) = y^2e^ysinx = C .

Answer 2

$\boxed {\textsf\red {Answer:}}$

$\underline{\textbf{Product rule of differentiation:}}$

$\boxed{\mathsf{\dfrac{d(uv)}{dx}=u\;\dfrac{dv}{dx}+v\;\dfrac{du}{dx}}}$

$\mathsf{Consider,}$

$\mathsf{(y\;cosx+1)dx+sinx\;dy=0}$

$\textsf{This can be written as}$

$\mathsf{y\;cosx+1+sinx\;\dfrac{dy}{dx}=0}$

$\mathsf{y\;cosx+sinx\;\dfrac{dy}{dx}=-1}$

$\textsf{Using Product rule, we get}$

$\mathsf{\dfrac{d(y\;sinx)}{dx}=-1}$

$\mathsf{d(y\;sinx)=-dx}d(ysinx)$

$\textsf{Integrating,}$

$\mathsf{\int\;d(y\;sinx)=-\int\;dx}$

$\mathsf{y\;sinx=-x+C}$

$\implies\boxed{\mathsf{x+(y\;sinx)=C}}$

$\textsf{which is the required solution. }$