Solve y=xsinx/1+x*operatorname{cosich}dy/x

$$y = \frac { x \sin x } { 1 + x } \cdot \operatorname { cosich } \frac { d y } { x }$$

Solve for d

$$\left\{\begin{matrix}d=\frac{4\left(x+1\right)e^{ch}}{-ie^{ix+2ch}+ie^{2ch-ix}+2\sin(x)}\text{, }&\left(-ie^{ix+2ch}+ie^{2ch-ix}+2\sin(x)\neq 0\text{ and }e^{ix+2ch}-e^{2ch-ix}+2i\sin(x)\neq 0\text{ and }x\neq -1\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }c=\frac{\pi \left(2in_{1}+i\right)}{2h}\right)\text{ or }\left(\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}\text{ and }x\neq -1\text{ and }h=0\right)\\d\in \mathrm{C}\text{, }&y=0\text{ and }x\neq 0\text{ and }x\neq -1\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x\left(x+1\right), the least common multiple of 1+x,x.

$$yx\left(x+1\right)=xx\sin(x)\cos(ich)\times \frac{dy}{x}$$

Use the distributive property to multiply yx by x+1.

$$yx^{2}+yx=xx\sin(x)\cos(ich)\times \frac{dy}{x}$$

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

$$yx^{2}+yx=x^{2}\sin(x)\cos(ich)\times \frac{dy}{x}$$

Express x^{2}\times \frac{dy}{x} as a single fraction.

$$yx^{2}+yx=\frac{x^{2}dy}{x}\sin(x)\cos(ich)$$

Cancel out x in both numerator and denominator.

$$yx^{2}+yx=dxy\sin(x)\cos(ich)$$

Swap sides so that all variable terms are on the left hand side.

$$dxy\sin(x)\cos(ich)=yx^{2}+yx$$

The equation is in standard form.

$$xy\sin(x)\cos(ich)d=xy+yx^{2}$$

Divide both sides by xy\sin(x)\cos(ich).

$$\frac{xy\sin(x)\cos(ich)d}{xy\sin(x)\cos(ich)}=\frac{xy\left(x+1\right)}{xy\sin(x)\cos(ich)}$$

Dividing by xy\sin(x)\cos(ich) undoes the multiplication by xy\sin(x)\cos(ich).

$$d=\frac{xy\left(x+1\right)}{xy\sin(x)\cos(ich)}$$

Divide yx\left(1+x\right) by xy\sin(x)\cos(ich).

$$d=\frac{x+1}{\sin(x)\cos(ich)}$$