Question

Find the value of \(\frac{{dy}}{{dx}}\) using equation 6.

\({e^y}\sin x = x + xy\)

Short answer

The value of \(\frac{{dy}}{{dx}}\) is\(\frac{{1 + y - {e^y}\cos x}}{{{e^y}\sin x - x}}\).

Step-by-step solution

Formula used

Equation 6: "\(\frac{{dy}}{{dx}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{dy}}}} = - \frac{{{F_x}}}{{{F_y}}}\), where\(F\)is the function of\(x\)and\({y^{\prime \prime }}\).

Find the value of \(\frac{{dy}}{{dx}}\)

As given equation,\({e^y}\sin x = x + xy\)

Let the function be, \(F(x,y) = {e^y}\sin x - x - xy \ldots \ldots \ldots (1)\)

Obtain the equation of \(\frac{{dy}}{{dx}}\) by using the equation 6,

\(\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}} \ldots \ldots \ldots (2)\)

Take the partial derivative with respect to \(x\) in the equation (1),

\(\begin{aligned}{l}\frac{{\partial F}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{e^y}\sin x - x - xy} \right)\\{F_x} = {e^y}(\cos x) - 1 - (1)y\\{F_x} = {e^y}\cos x - 1 - y\end{aligned}\)

Thus, the partial derivate of\(x\)is\({F_x} = {e^y}\cos x - 1 - y\).

Take the partial derivative with respect to \(y\) in the equation (1),

\(\begin{aligned}{l}\frac{{\partial F}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{e^y}\sin x - x - xy} \right)\\{F_y} = {e^y}(\sin x) - 0 - x(1)\\{F_y} = {e^y}\sin x - x\end{aligned}\)

Thus, the partial derivate of\(y\)is\({F_y} = {e^y}\sin x - x\).

Substitute the respective values in the equation (2),

\(\begin{aligned}{l}\frac{{dy}}{{dx}} = - \frac{{{F_x}}}{{{F_y}}}\\\frac{{dy}}{{dx}} = - \frac{{{e^y}\cos x - 1 - y}}{{{e^y}\sin x - x}}\\\frac{{dy}}{{dx}} = \frac{{1 + y - {e^y}\cos x}}{{{e^y}\sin x - x}}\end{aligned}\)

Therefore, the value of \(\frac{{dy}}{{dx}}\) is\(\frac{{1 + y - {e^y}\cos x}}{{{e^y}\sin x - x}}\).