Solved example of common monomial factor
Rewrite the differential equation
Factoring by $y$
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
Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
Solve the integral $\int\frac{1}{y}dy$ and replace the result in the differential equation
Expand the fraction $\frac{1-x}{x^2}$ into $2$ simpler fractions with common denominator $x^2$
Simplify the resulting fractions
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
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$
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 $-2$
Simplify the expression inside the integral
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
Solve the integral $\int\frac{1-x}{x^2}dx$ and replace the result in the differential equation
Take the variable outside of the logarithm
Simplifying the logarithm
Simplify $e^{\left(\frac{1}{-x}-\ln\left(x\right)+C_0\right)}$ by applying the properties of exponents and logarithms
Simplify $e^{\left(\frac{1}{-x}+C_0\right)}$ applying properties of exponents
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
Any expression to the power of $1$ is equal to that same expression
Multiply the fraction and term
Find the explicit solution to the differential equation. We need to isolate the variable $y$
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