Solve +(e^x+1)sinydy=0

$$+ ( e ^ { x } + 1 ) \sin y d y = 0$$

Solve for x

$$x\in \mathrm{R}$$

$$y=0\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }d=\frac{\pi n_{1}}{y^{2}}$$

Steps Using Definition of Logarithm

Multiply y and y to get y^{2}.

$$\left(e^{x}+1\right)\sin(y^{2}d)=0$$

Use the distributive property to multiply e^{x}+1 by \sin(y^{2}d).

$$e^{x}\sin(y^{2}d)+\sin(y^{2}d)=0$$

Use the rules of exponents and logarithms to solve the equation.

$$\sin(dy^{2})e^{x}+\sin(dy^{2})=0$$

Subtract \sin(y^{2}d) from both sides of the equation.

$$\sin(dy^{2})e^{x}=-\sin(dy^{2})$$

Divide both sides by \sin(y^{2}d).

$$e^{x}=-1$$

Take the logarithm of both sides of the equation.

$$\log(e^{x})=\log(-1)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$x\log(e)=\log(-1)$$

Divide both sides by \log(e).

$$x=\frac{\log(-1)}{\log(e)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x=\log_{e}\left(-1\right)$$

Solve for d

$$\left\{\begin{matrix}d=\frac{\pi n_{1}}{y^{2}}\text{, }n_{1}\in \mathrm{Z}\text{, }&y\neq 0\\d\in \mathrm{R}\text{, }&y=0\end{matrix}\right.$$