( \frac { y ^ { 2 } } { ( y - x ) ^ { 2 } } - \frac { 1 } { x } ) d x + ( \frac { 1 } { y } - \frac { x ^ { 2 } } { ( x - y ) ^ { 2 } } ) d y = 0
Solve for d (complex solution)
d=0
y\neq x\text{ and }x\neq 0\text{ and }y\neq 0
Solve for d
d=0
x\neq 0\text{ and }y\neq 0\text{ and }y\neq x
Solve for x (complex solution)
x\in \mathrm{C}\setminus y,0
d=0\text{ and }y\neq 0
Solve for x
x\in \mathrm{R}\setminus 0,y
d=0\text{ and }y\neq 0
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\left(\frac{y^{2}}{\left(y-x\right)^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Multiply both sides of the equation by xy\left(-x+y\right)^{2}, the least common multiple of \left(y-x\right)^{2},x,y,\left(x-y\right)^{2}.
\left(\frac{y^{2}}{y^{2}-2yx+x^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-x\right)^{2}.
\left(\frac{y^{2}}{\left(-x+y\right)^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Factor y^{2}-2yx+x^{2}.
\left(\frac{y^{2}x}{x\left(-x+y\right)^{2}}-\frac{\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-x+y\right)^{2} and x is x\left(-x+y\right)^{2}. Multiply \frac{y^{2}}{\left(-x+y\right)^{2}} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{\left(-x+y\right)^{2}}{\left(-x+y\right)^{2}}.
\frac{y^{2}x-\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}}dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Since \frac{y^{2}x}{x\left(-x+y\right)^{2}} and \frac{\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}} have the same denominator, subtract them by subtracting their numerators.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Do the multiplications in y^{2}x-\left(-x+y\right)^{2}.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dx^{2}y\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Multiply x and x to get x^{2}.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dx^{2}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+y\right)^{2}.
\frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)d}{x\left(-x+y\right)^{2}}x^{2}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}d as a single fraction.
\frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)dx^{2}}{x\left(-x+y\right)^{2}}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)d}{x\left(-x+y\right)^{2}}x^{2} as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)}{\left(-x+y\right)^{2}}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Cancel out x in both numerator and denominator.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)}{\left(-x+y\right)^{2}}y as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right) as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Multiply y and y to get y^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{x^{2}-2xy+y^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(-x+y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Factor x^{2}-2xy+y^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{\left(-x+y\right)^{2}}{y\left(-x+y\right)^{2}}-\frac{x^{2}y}{y\left(-x+y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y and \left(-x+y\right)^{2} is y\left(-x+y\right)^{2}. Multiply \frac{1}{y} times \frac{\left(-x+y\right)^{2}}{\left(-x+y\right)^{2}}. Multiply \frac{x^{2}}{\left(-x+y\right)^{2}} times \frac{y}{y}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(-x+y\right)^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(-x+y\right)^{2}=0
Since \frac{\left(-x+y\right)^{2}}{y\left(-x+y\right)^{2}} and \frac{x^{2}y}{y\left(-x+y\right)^{2}} have the same denominator, subtract them by subtracting their numerators.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(-x+y\right)^{2}=0
Do the multiplications in \left(-x+y\right)^{2}-x^{2}y.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(x^{2}-2xy+y^{2}\right)=0
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+y\right)^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)d}{y\left(-x+y\right)^{2}}y^{2}x\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}d as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)dy^{2}}{y\left(-x+y\right)^{2}}x\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)d}{y\left(-x+y\right)^{2}}y^{2} as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)}{\left(-x+y\right)^{2}}x\left(x^{2}-2xy+y^{2}\right)=0
Cancel out y in both numerator and denominator.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)}{\left(-x+y\right)^{2}}x as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}=0
Express \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right) as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)+dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}=0
Since \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}} and \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}} have the same denominator, add them by adding their numerators.
\frac{-ydx^{5}+2y^{2}dx^{4}-y^{3}dx^{3}+y^{3}dx^{4}-2y^{4}dx^{3}+y^{5}dx^{2}+2y^{2}dx^{4}-4dx^{3}y^{3}+2y^{4}dx^{2}-y^{3}dx^{3}+2y^{4}dx^{2}-y^{5}dx+ydx^{5}-2y^{2}dx^{4}+y^{3}dx^{3}-2y^{2}dx^{4}+4dy^{3}x^{3}-2y^{4}dx^{2}+y^{3}dx^{3}-2y^{4}dx^{2}+y^{5}dx-y^{2}dx^{5}+2dy^{3}x^{4}-y^{4}dx^{3}}{\left(-x+y\right)^{2}}=0
Do the multiplications in dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)+dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right).
\frac{-y^{2}dx^{5}+y^{5}dx^{2}-3y^{4}dx^{3}+3y^{3}dx^{4}}{\left(-x+y\right)^{2}}=0
Combine like terms in -ydx^{5}+2y^{2}dx^{4}-y^{3}dx^{3}+y^{3}dx^{4}-2y^{4}dx^{3}+y^{5}dx^{2}+2y^{2}dx^{4}-4dx^{3}y^{3}+2y^{4}dx^{2}-y^{3}dx^{3}+2y^{4}dx^{2}-y^{5}dx+ydx^{5}-2y^{2}dx^{4}+y^{3}dx^{3}-2y^{2}dx^{4}+4dy^{3}x^{3}-2y^{4}dx^{2}+y^{3}dx^{3}-2y^{4}dx^{2}+y^{5}dx-y^{2}dx^{5}+2dy^{3}x^{4}-y^{4}dx^{3}.
\frac{d\left(-x+y\right)x^{2}y^{2}\left(x-y\right)^{2}}{\left(-x+y\right)^{2}}=0
Factor the expressions that are not already factored in \frac{-y^{2}dx^{5}+y^{5}dx^{2}-3y^{4}dx^{3}+3y^{3}dx^{4}}{\left(-x+y\right)^{2}}.
\frac{dx^{2}y^{2}\left(x-y\right)^{2}}{-x+y}=0
Cancel out -x+y in both numerator and denominator.
\frac{dx^{2}y^{2}\left(x^{2}-2xy+y^{2}\right)}{-x+y}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.
dx^{2}y^{2}\left(x^{2}-2xy+y^{2}\right)=0
Multiply both sides of the equation by -x+y.
dy^{2}x^{4}-2dx^{3}y^{3}+dx^{2}y^{4}=0
Use the distributive property to multiply dx^{2}y^{2} by x^{2}-2xy+y^{2}.
\left(y^{2}x^{4}-2x^{3}y^{3}+x^{2}y^{4}\right)d=0
Combine all terms containing d.
\left(y^{2}x^{4}+x^{2}y^{4}-2x^{3}y^{3}\right)d=0
The equation is in standard form.
d=0
Divide 0 by y^{2}x^{4}-2x^{3}y^{3}+x^{2}y^{4}.
\left(\frac{y^{2}}{\left(y-x\right)^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Multiply both sides of the equation by xy\left(-x+y\right)^{2}, the least common multiple of \left(y-x\right)^{2},x,y,\left(x-y\right)^{2}.
\left(\frac{y^{2}}{y^{2}-2yx+x^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-x\right)^{2}.
\left(\frac{y^{2}}{\left(-x+y\right)^{2}}-\frac{1}{x}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Factor y^{2}-2yx+x^{2}.
\left(\frac{y^{2}x}{x\left(-x+y\right)^{2}}-\frac{\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}}\right)dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-x+y\right)^{2} and x is x\left(-x+y\right)^{2}. Multiply \frac{y^{2}}{\left(-x+y\right)^{2}} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{\left(-x+y\right)^{2}}{\left(-x+y\right)^{2}}.
\frac{y^{2}x-\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}}dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Since \frac{y^{2}x}{x\left(-x+y\right)^{2}} and \frac{\left(-x+y\right)^{2}}{x\left(-x+y\right)^{2}} have the same denominator, subtract them by subtracting their numerators.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dxxy\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Do the multiplications in y^{2}x-\left(-x+y\right)^{2}.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dx^{2}y\left(-x+y\right)^{2}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Multiply x and x to get x^{2}.
\frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}dx^{2}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+y\right)^{2}.
\frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)d}{x\left(-x+y\right)^{2}}x^{2}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{y^{2}x-x^{2}+2xy-y^{2}}{x\left(-x+y\right)^{2}}d as a single fraction.
\frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)dx^{2}}{x\left(-x+y\right)^{2}}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{\left(y^{2}x-x^{2}+2xy-y^{2}\right)d}{x\left(-x+y\right)^{2}}x^{2} as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)}{\left(-x+y\right)^{2}}y\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Cancel out x in both numerator and denominator.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)}{\left(-x+y\right)^{2}}y as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dyxy\left(-x+y\right)^{2}=0
Express \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right) as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(x-y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Multiply y and y to get y^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{x^{2}-2xy+y^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{1}{y}-\frac{x^{2}}{\left(-x+y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
Factor x^{2}-2xy+y^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\left(\frac{\left(-x+y\right)^{2}}{y\left(-x+y\right)^{2}}-\frac{x^{2}y}{y\left(-x+y\right)^{2}}\right)dy^{2}x\left(-x+y\right)^{2}=0
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of y and \left(-x+y\right)^{2} is y\left(-x+y\right)^{2}. Multiply \frac{1}{y} times \frac{\left(-x+y\right)^{2}}{\left(-x+y\right)^{2}}. Multiply \frac{x^{2}}{\left(-x+y\right)^{2}} times \frac{y}{y}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(-x+y\right)^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(-x+y\right)^{2}=0
Since \frac{\left(-x+y\right)^{2}}{y\left(-x+y\right)^{2}} and \frac{x^{2}y}{y\left(-x+y\right)^{2}} have the same denominator, subtract them by subtracting their numerators.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(-x+y\right)^{2}=0
Do the multiplications in \left(-x+y\right)^{2}-x^{2}y.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}dy^{2}x\left(x^{2}-2xy+y^{2}\right)=0
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+y\right)^{2}.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)d}{y\left(-x+y\right)^{2}}y^{2}x\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{x^{2}-2xy+y^{2}-x^{2}y}{y\left(-x+y\right)^{2}}d as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)dy^{2}}{y\left(-x+y\right)^{2}}x\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{\left(x^{2}-2xy+y^{2}-x^{2}y\right)d}{y\left(-x+y\right)^{2}}y^{2} as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)}{\left(-x+y\right)^{2}}x\left(x^{2}-2xy+y^{2}\right)=0
Cancel out y in both numerator and denominator.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right)=0
Express \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)}{\left(-x+y\right)^{2}}x as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}+\frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}=0
Express \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x}{\left(-x+y\right)^{2}}\left(x^{2}-2xy+y^{2}\right) as a single fraction.
\frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)+dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}}=0
Since \frac{dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}} and \frac{dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right)}{\left(-x+y\right)^{2}} have the same denominator, add them by adding their numerators.
\frac{-ydx^{5}+2y^{2}dx^{4}-y^{3}dx^{3}+y^{3}dx^{4}-2y^{4}dx^{3}+y^{5}dx^{2}+2y^{2}dx^{4}-4dx^{3}y^{3}+2y^{4}dx^{2}-y^{3}dx^{3}+2y^{4}dx^{2}-y^{5}dx+ydx^{5}-2y^{2}dx^{4}+y^{3}dx^{3}-2y^{2}dx^{4}+4dy^{3}x^{3}-2y^{4}dx^{2}+y^{3}dx^{3}-2y^{4}dx^{2}+y^{5}dx-y^{2}dx^{5}+2dy^{3}x^{4}-y^{4}dx^{3}}{\left(-x+y\right)^{2}}=0
Do the multiplications in dx\left(-x^{2}+xy^{2}+2xy-y^{2}\right)y\left(x^{2}-2xy+y^{2}\right)+dy\left(x^{2}-2xy+y^{2}-yx^{2}\right)x\left(x^{2}-2xy+y^{2}\right).
\frac{-y^{2}dx^{5}+y^{5}dx^{2}-3y^{4}dx^{3}+3y^{3}dx^{4}}{\left(-x+y\right)^{2}}=0
Combine like terms in -ydx^{5}+2y^{2}dx^{4}-y^{3}dx^{3}+y^{3}dx^{4}-2y^{4}dx^{3}+y^{5}dx^{2}+2y^{2}dx^{4}-4dx^{3}y^{3}+2y^{4}dx^{2}-y^{3}dx^{3}+2y^{4}dx^{2}-y^{5}dx+ydx^{5}-2y^{2}dx^{4}+y^{3}dx^{3}-2y^{2}dx^{4}+4dy^{3}x^{3}-2y^{4}dx^{2}+y^{3}dx^{3}-2y^{4}dx^{2}+y^{5}dx-y^{2}dx^{5}+2dy^{3}x^{4}-y^{4}dx^{3}.
\frac{d\left(-x+y\right)x^{2}y^{2}\left(x-y\right)^{2}}{\left(-x+y\right)^{2}}=0
Factor the expressions that are not already factored in \frac{-y^{2}dx^{5}+y^{5}dx^{2}-3y^{4}dx^{3}+3y^{3}dx^{4}}{\left(-x+y\right)^{2}}.
\frac{dx^{2}y^{2}\left(x-y\right)^{2}}{-x+y}=0
Cancel out -x+y in both numerator and denominator.
\frac{dx^{2}y^{2}\left(x^{2}-2xy+y^{2}\right)}{-x+y}=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.
dx^{2}y^{2}\left(x^{2}-2xy+y^{2}\right)=0
Multiply both sides of the equation by -x+y.
dy^{2}x^{4}-2dx^{3}y^{3}+dx^{2}y^{4}=0
Use the distributive property to multiply dx^{2}y^{2} by x^{2}-2xy+y^{2}.
\left(y^{2}x^{4}-2x^{3}y^{3}+x^{2}y^{4}\right)d=0
Combine all terms containing d.
\left(y^{2}x^{4}+x^{2}y^{4}-2x^{3}y^{3}\right)d=0
The equation is in standard form.
d=0
Divide 0 by y^{2}x^{4}-2x^{3}y^{3}+x^{2}y^{4}.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}