Solve y=(k+1)x^k^2-k-3dx

$$y = ( k + 1 ) x ^ { k ^ { 2 } - k - 3 } d x$$

Solve for d (complex solution)

$$\left\{\begin{matrix}d=\frac{y}{\left(k+1\right)x^{\left(k-2\right)\left(k+1\right)}}\text{, }&\left(x\neq 0\text{ and }k\neq -1\right)\text{ or }k=2\\d\in \mathrm{C}\text{, }&\left(k=-1\text{ or }x=0\right)\text{ and }k\neq 2\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply k+1 by x^{k^{2}-k-3}.

$$y=\left(kx^{k^{2}-k-3}+x^{k^{2}-k-3}\right)dx$$

Use the distributive property to multiply kx^{k^{2}-k-3}+x^{k^{2}-k-3} by d.

$$y=\left(kx^{k^{2}-k-3}d+x^{k^{2}-k-3}d\right)x$$

Use the distributive property to multiply kx^{k^{2}-k-3}d+x^{k^{2}-k-3}d by x.

$$y=kx^{k^{2}-k-3}dx+x^{k^{2}-k-3}dx$$

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

$$kx^{k^{2}-k-3}dx+x^{k^{2}-k-3}dx=y$$

Combine all terms containing d.

$$\left(kx^{k^{2}-k-3}x+x^{k^{2}-k-3}x\right)d=y$$

The equation is in standard form.

$$\left(kx^{k^{2}-k-2}+x^{k^{2}-k-2}\right)d=y$$

Divide both sides by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$\frac{\left(kx^{k^{2}-k-2}+x^{k^{2}-k-2}\right)d}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}=\frac{y}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}$$

Dividing by kx^{k^{2}-k-2}+x^{k^{2}-k-2} undoes the multiplication by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$d=\frac{y}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}$$

Divide y by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$d=\frac{y}{\left(k+1\right)x^{\left(k-2\right)\left(k+1\right)}}$$

Solve for d

$$\left\{\begin{matrix}d=\frac{y}{\left(k+1\right)x^{\left(k-2\right)\left(k+1\right)}}\text{, }&\left(k\neq -1\text{ and }x>0\right)\text{ or }\left(k\neq -1\text{ and }Denominator(\left(k-2\right)\left(k+1\right))\text{bmod}2=1\text{ and }x<0\text{ and }Denominator(k^{2}-k)\text{bmod}2=1\right)\\d\in \mathrm{R}\text{, }&\left(y=0\text{ and }x=0\text{ and }k<\frac{1-\sqrt{13}}{2}\right)\text{ or }\left(y=0\text{ and }x=0\text{ and }k>\frac{\sqrt{13}+1}{2}\right)\text{ or }\left(y=0\text{ and }k=-1\text{ and }x\neq 0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply k+1 by x^{k^{2}-k-3}.

$$y=\left(kx^{k^{2}-k-3}+x^{k^{2}-k-3}\right)dx$$

Use the distributive property to multiply kx^{k^{2}-k-3}+x^{k^{2}-k-3} by d.

$$y=\left(kx^{k^{2}-k-3}d+x^{k^{2}-k-3}d\right)x$$

Use the distributive property to multiply kx^{k^{2}-k-3}d+x^{k^{2}-k-3}d by x.

$$y=kx^{k^{2}-k-3}dx+x^{k^{2}-k-3}dx$$

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

$$kx^{k^{2}-k-3}dx+x^{k^{2}-k-3}dx=y$$

Combine all terms containing d.

$$\left(kx^{k^{2}-k-3}x+x^{k^{2}-k-3}x\right)d=y$$

The equation is in standard form.

$$\left(kx^{k^{2}-k-2}+x^{k^{2}-k-2}\right)d=y$$

Divide both sides by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$\frac{\left(kx^{k^{2}-k-2}+x^{k^{2}-k-2}\right)d}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}=\frac{y}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}$$

Dividing by kx^{k^{2}-k-2}+x^{k^{2}-k-2} undoes the multiplication by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$d=\frac{y}{kx^{k^{2}-k-2}+x^{k^{2}-k-2}}$$

Divide y by kx^{k^{2}-k-2}+x^{k^{2}-k-2}.

$$d=\frac{y}{\left(k+1\right)x^{\left(k-2\right)\left(k+1\right)}}$$