Question

Solve the differential equation $ydx-\left( x+2{{y}^{2}} \right)dy=0$

Answer 1

Hint: Manipulate the given equation to get the value of $\dfrac{dy}{dx}$ from the given equation. Try to convert the equation into integrating factor type of equation. Use the formula of integrating factor to find the original function. Now write the original function, which is the required result in this question.

Complete step-by-step answer:

Given differential equation in the question is written in form of

$ydx-\left( x+2{{y}^{2}} \right)dy=0$

Multiply dy inside to remove the bracket, we get it as

$ydx-xdy-2{{y}^{2}}dy=0$

By adding $xdy+2{{y}^{2}}dy$ on both sides, we get the equation as

$ydx=xdy+2{{y}^{2}}dy$

By taking dy as common, we can write the equation in form of

$ydx=\left( x+2{{y}^{2}} \right)dy$

By dividing with dx on both sides, we can write it as

$y=\left( x+2{{y}^{2}} \right)\dfrac{dy}{dx}$

By dividing with $\left( x+2{{y}^{2}} \right)$ on both sides, we get it in form of

$\dfrac{dy}{dx}=\dfrac{y}{x+2{{y}^{2}}}...................(1)$

It is not in the form of $\dfrac{dy}{dx}+Py=Q$ . So, we find an equation for $\dfrac{dx}{dy}$ and check it is in this required form.

By reversing the equation (1), we get an equation given by

$\dfrac{dx}{dy}=\dfrac{x+2{{y}^{2}}}{y}$

By writing the fraction in two term, we have formula as

$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$

By substituting this in right hand side, we can write equation as

$\dfrac{dx}{dy}=\dfrac{x}{y}+\dfrac{2{{y}^{2}}}{y}$

By simplifying the above term, we can write the equation as

$\dfrac{dx}{dy}=\dfrac{x}{y}+2y$

By subtracting $\dfrac{x}{y}$ on both sides, we can write the equation as

\[\dfrac{dx}{dy}-\dfrac{x}{y}=\dfrac{x}{y}-\dfrac{x}{y}+2y\]

By simplifying the equation, we should write it in form of

$\dfrac{dx}{dy}-\dfrac{x}{y}=2y$

By comparing it to the equation, we can write it in the form of

$\dfrac{dx}{dy}+{{P}_{1}}x={{Q}_{1}}$

We can get value of variables as given in terms of y as

${{P}_{1}}=\dfrac{-1}{y},{{Q}_{1}}=2y$

Now we know the formula of integrating factor as given by

$IF={{e}^{\int{{{P}_{1}}dy}}}$

By substituting their values in the equation, we can write it as

$IF={{e}^{\int{-\dfrac{1}{y}dy}}}$

By simplifying the integration in the equation, we can write it as

$IF={{e}^{-\log y}}$

By basic knowledge of exponential, we know the formula

$a\log b=\log {{b}^{a}}$

We can send -1 inside the logarithm, we can write it as

$IF={{e}^{\log \dfrac{1}{y}}}$

By cancelling e into the term ${{e}^{\log k}}=k$ we can write it as

$IF=\dfrac{1}{y}$

After getting integrating factor, we have the formula given as

$x.IF=\int{\left( Q\times IF \right)dy+c}$

By substituting their values, we can write the equation the equation as

$x\times \dfrac{1}{y}=\int{2y\times \dfrac{1}{y}dy+c}$

By simplifying above terms, we can write the equation in form of

$\dfrac{x}{y}=\int{2dy+c}$

By using basic formula in integration, which is given by

$\int{kdy=ky+c}$ , where k is constant.

So on solving $\int{2dy+c}$, we get $2y+c$

Then we can write the equation as

$\dfrac{x}{y}=2y+c$

By multiply with y on both sides, we get the equation as

x = y (2y + c)

By simplifying we can write the value of x in form of

$x=2{{y}^{2}}+cy$

Therefore this is the solution of the given differential equation.

Hence, general solution of differential equation $ydx-\left( x+2{{y}^{2}} \right)dy=0$ is $x=2{{y}^{2}}+cy$, where c is constant.

Note: Be careful with integrating factor formula, because the integrating factor is not a final function it just makes a formula for final solution. Be careful with the places of x, y after substituting. As the equation is $\dfrac{dx}{dy}$ so, the formula will have x on the left hand side. The idea of using the $\dfrac{dx}{dy}$ is very crucial. Do it carefully.