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Common monomial factor Calculator

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1

Solved example of common monomial factor

$\frac{dx}{dy}=y+xy$
2

Factoring by $y$

$\frac{dx}{dy}=y\left(1+x\right)$
3

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

$\frac{1}{1+x}dx=y\cdot dy$
4

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

$\int\frac{1}{1+x}dx=\int ydy$

We can solve the integral $\int\frac{1}{1+x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a part of the integral with a new variable, which when substituted makes the integral easier. We see that $1+x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=1+x$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=dx$

Substituting $u$ and $dx$ in the integral and simplify

$\int\frac{1}{u}du$

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

$\ln\left(u\right)$

Substitute $u$ back with the value that we assigned to it: $1+x$

$\ln\left(1+x\right)$
5

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

$\ln\left(1+x\right)=\int ydy$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{1}{2}y^2$
6

Solve the integral $\int ydy$ and replace the result in the differential equation

$\ln\left(1+x\right)=\frac{1}{2}y^2$
7

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

$\ln\left(1+x\right)=\frac{1}{2}y^2+C_0$
8

Take the variable outside of the logarithm

$e^{\ln\left(1+x\right)}=e^{\left(\frac{1}{2}y^2+C_0\right)}$
9

Simplifying the logarithm

$1+x=e^{\left(\frac{1}{2}y^2+C_0\right)}$
10

Simplify $e^{\left(\frac{1}{2}y^2+C_0\right)}$ applying properties of exponents

$1+x=C_0e^{\frac{1}{2}y^2}$
11

We need to isolate the dependent variable $x$, we can do that by subtracting $1$ from both sides of the equation

$x=C_0e^{\frac{1}{2}y^2}-1$

Final Answer

$x=C_0e^{\frac{1}{2}y^2}-1$

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