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# Common Monomial Factor Calculator

## Get detailed solutions to your math problems with our Common Monomial Factor step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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###  Difficult Problems

1

Solved example of common monomial factor

$x^2\frac{dy}{dx}=y-xy$
2

Rewrite the differential equation

$\frac{dy}{dx}=\frac{y-xy}{x^2}$
3

Factoring by $y$

$\frac{dy}{dx}=\frac{y\left(1-x\right)}{x^2}$

4

Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side of the equality

$\frac{1}{y}dy=\frac{1-x}{x^2}dx$
5

Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$

$\int\frac{1}{y}dy=\int\frac{1-x}{x^2}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\ln\left(y\right)$
6

Solve the integral $\int\frac{1}{y}dy$ and replace the result in the differential equation

$\ln\left(y\right)=\int\frac{1-x}{x^2}dx$

Expand the fraction $\frac{1-x}{x^2}$ into $2$ simpler fractions with common denominator $x^2$

$\int\left(\frac{1}{x^2}+\frac{-x}{x^2}\right)dx$

Simplify the resulting fractions

$\int\left(\frac{1}{x^2}+\frac{-1}{x}\right)dx$

Expand the integral $\int\left(\frac{1}{x^2}+\frac{-1}{x}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$\int\frac{1}{x^2}dx+\int\frac{-1}{x}dx$

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\int\frac{1}{x^2}dx-\ln\left(x\right)$

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int x^{-2}dx-\ln\left(x\right)$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $-2$

$\frac{x^{-1}}{-1}-\ln\left(x\right)$

Simplify the expression inside the integral

$\frac{1}{-x}-\ln\left(x\right)$

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{-x}-\ln\left(x\right)+C_0$
7

Solve the integral $\int\frac{1-x}{x^2}dx$ and replace the result in the differential equation

$\ln\left(y\right)=\frac{1}{-x}-\ln\left(x\right)+C_0$

Take the variable outside of the logarithm

$e^{\ln\left(y\right)}=e^{\left(\frac{1}{-x}-\ln\left(x\right)+C_0\right)}$

Simplifying the logarithm

$y=e^{\left(\frac{1}{-x}-\ln\left(x\right)+C_0\right)}$

Simplify $e^{\left(\frac{1}{-x}-\ln\left(x\right)+C_0\right)}$ by applying the properties of exponents and logarithms

$y=x^{-1}e^{\left(\frac{1}{-x}+C_0\right)}$

Simplify $e^{\left(\frac{1}{-x}+C_0\right)}$ applying properties of exponents

$y=C_1x^{-1}e^{\frac{1}{-x}}$

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$y=C_1\frac{1}{x^{1}}e^{\frac{1}{-x}}$

Any expression to the power of $1$ is equal to that same expression

$y=C_1\frac{1}{x}e^{\frac{1}{-x}}$

Multiply the fraction and term

$y=\frac{C_1e^{\frac{1}{-x}}}{x}$
8

Find the explicit solution to the differential equation. We need to isolate the variable $y$

$y=\frac{C_1e^{\frac{1}{-x}}}{x}$

$y=\frac{C_1e^{\frac{1}{-x}}}{x}$