Question

How do you integrate $ydx=2\left( x+y \right)dy$?

Answer 1

Divide both the sides with $ydy$ and take the terms of x to the L.H.S to form a linear differential equation in x of the form $\dfrac{dx}{dy}+Px=Q$, where P and Q may be constants or functions of y. Now, find the integrating factor by using the formula $I.F={{e}^{\int{Pdy}}}$ and write the solution of the differential equation as: $x\times I.F=\int{\left( I.F\times Q \right)}dy$. Perform the required integral and add the constant to integration (c) in the end to get the solution.

Complete step by step solution:
Here we have been provided with the differential equation $ydx=2\left( x+y \right)dy$ and we are asked to integrate it. That means we need to find its solution. Let us try to form a linear differential equation.
$\because ydx=2\left( x+y \right)dy$
Dividing both the sides with $ydy$ we get,
$\begin{aligned} & \Rightarrow \dfrac{dx}{dy}=\dfrac{2\left( x+y \right)}{y} \\\ & \Rightarrow \dfrac{dx}{dy}=\dfrac{2x}{y}+2 \\\ \end{aligned}$
Taking the terms of x to the L.H.S we have,
$\begin{aligned} & \Rightarrow \dfrac{dx}{dy}=\dfrac{2\left( x+y \right)}{y} \\\ & \Rightarrow \dfrac{dx}{dy}-\dfrac{2x}{y}=2 \\\ & \Rightarrow \dfrac{dx}{dy}+\left( -\dfrac{2}{y} \right)x=2 \\\ \end{aligned}$
Clearly, the above differential equation is of the form $\dfrac{dx}{dy}+Px=Q$, where $P=\left( -\dfrac{2}{y} \right)$ and $Q=2$. This form of the differential equation is called linear differential equation in x. So, to find the solution first we need to find the integrating factor (I.F) given by the formula: $I.F={{e}^{\int{Pdy}}}$.
$\begin{aligned} & \Rightarrow I.F={{e}^{\int{\left( -\dfrac{2}{y} \right)dy}}} \\\ & \Rightarrow I.F={{e}^{-2\int{\dfrac{1}{y}dy}}} \\\ \end{aligned}$
We know that $\int{\dfrac{dy}{y}}=\ln y$, so we have,
$\Rightarrow I.F={{e}^{-2\ln y}}$
Using the property of log given as $m\log a=\log {{a}^{m}}$ we get,
$\begin{aligned} & \Rightarrow I.F={{e}^{\ln \left( {{y}^{-2}} \right)}} \\\ & \Rightarrow I.F={{y}^{-2}} \\\ & \Rightarrow I.F=\dfrac{1}{{{y}^{2}}} \\\ \end{aligned}$
Now, we know that the solution of a linear differential equation in x is given as $x\times I.F=\int{\left( I.F\times Q \right)}dy$, so we have,
$\begin{aligned} & \Rightarrow x\times \dfrac{1}{{{y}^{2}}}=\int{\left( \dfrac{1}{{{y}^{2}}}\times 2 \right)}dy \\\ & \Rightarrow \dfrac{x}{{{y}^{2}}}=2\int{\left( \dfrac{1}{{{y}^{2}}} \right)}dy \\\ \end{aligned}$
We know that $\int{\dfrac{1}{{{y}^{2}}}dy}=\left( -\dfrac{1}{y} \right)$, so we have,
$\begin{aligned} & \Rightarrow \dfrac{x}{{{y}^{2}}}=2\left( -\dfrac{1}{y} \right)+c \\\ & \Rightarrow \dfrac{x}{{{y}^{2}}}+\dfrac{2}{y}=c \\\ & \Rightarrow \dfrac{x+2y}{{{y}^{2}}}=c \\\ \end{aligned}$
Here ‘c’ is the constant of integration.
Here ‘c’ is the constant of integration.
Hence, the above relation is our answer.

Note: There can be another method also to solve the above question. We can also form a homogeneous differential equation and by using the substitution $x=vy$ we can simplify the given equation. In the end the solution may look different but on simplifying it to a certain step we will get the same answer. Here you cannot form a linear differential equation in y of the form $\dfrac{dy}{dx}+Py=Q$ so you need to be careful at the initial step of the solution.