# The integrating factor of the differential equation $\left( 1-{{y}^{2}} \right)\dfrac{dx}{dy}+yx=ay$ is

[a] $\dfrac{1}{{{y}^{2}}-1}$

[b] $\dfrac{1}{\sqrt{{{y}^{2}}-1}}$

[c] $\dfrac{1}{1-{{y}^{2}}}$

[d] $\dfrac{1}{\sqrt{1-{{y}^{2}}}}$

Answer

Verified

380.4k+ views

Hint: Integrating factor of a differential equation is a term with which we should multiply the differential equation so that it becomes exact. An exact differential equation is the differential equation $Mdx+Ndy=0$ which satisfies the Euler criterion for exactness, i.e. $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$. In a linear order differential equation, i.e. equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ the integrating factor $IF={{e}^{\int{P\left( x \right)dx}}}$. Convert the above differential equation in the exact form by dividing on both sides by $1-{{y}^{2}}$ and find the integrating factor using the above formula for IF.

Complete step by step solution:

We have $\left( 1-{{y}^{2}} \right)\dfrac{dx}{dy}+yx=ay$

Dividing both sides by $1-{{y}^{2}}$, we get

$\dfrac{1-{{y}^{2}}}{1-{{y}^{2}}}\dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$

$\Rightarrow \dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$, which is of the form $\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$, where $P\left( y \right)=\dfrac{y}{1-{{y}^{2}}}$ and $Q\left( y \right)=\dfrac{ay}{1-{{y}^{2}}}$

We have Integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}$.

Let $I=\int{P\left( y \right)dy}$

So, we have

$I=\int{\dfrac{y}{1-{{y}^{2}}}dy}$

Put $1-{{y}^{2}}=z$

Differentiating both sides, we get

\[\begin{align}

& -2ydy=dz \\

& \Rightarrow ydy=-\dfrac{dz}{2} \\

\end{align}\]

So, we have

\[\begin{align}

& I=\int{\dfrac{-dz}{2z}} \\

& =-\dfrac{1}{2}\int{\dfrac{dz}{z}} \\

\end{align}\]

We know that $\int{\dfrac{dx}{x}=\ln x+c}$

Using, we get

$I=-\dfrac{1}{2}\ln z$

Returning to the original variable, we get

$\begin{align}

& I=-\dfrac{1}{2}\ln \left( \left| 1-{{y}^{2}} \right| \right) \\

& \Rightarrow I=\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right) \\

\end{align}$

Hence the integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}={{e}^{I}}={{e}^{\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right)}}$

We know that ${{e}^{\ln x}}=x$

Using we get

$IF=\dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}}$.

Hence, options [b] and [d] are correct.

Note: [1]A differential equation when in exact form can be written in the form $du=dv$.

In a Linear differential equation when multiplied by Integrating factor we have $u=y\cdot IF$ and $v=\int{Q\left( x \right)\cdot IFdx}$.

[2] Euler’s criterion for exactness is a direct result of the fact $\dfrac{\partial f}{\partial x\partial y}=\dfrac{\partial f}{\partial y\partial x}$ .

[3] Sometimes, the following identities help in converting a differential equation in the exact form:

[a] $xdy+ydx=d(xy)$

[b] $dx+dy=d(x+y)$

[c] $\dfrac{xdy-ydx}{{{x}^{2}}}=d\left( \dfrac{y}{x} \right)$

[d] $\dfrac{dx}{x}=d\left( \ln x \right)$

[e] $m{{x}^{m-1}}{{y}^{n}}+n{{x}^{m}}{{y}^{n-1}}=d\left( {{x}^{m}}{{y}^{n}} \right)$

Complete step by step solution:

We have $\left( 1-{{y}^{2}} \right)\dfrac{dx}{dy}+yx=ay$

Dividing both sides by $1-{{y}^{2}}$, we get

$\dfrac{1-{{y}^{2}}}{1-{{y}^{2}}}\dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$

$\Rightarrow \dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$, which is of the form $\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$, where $P\left( y \right)=\dfrac{y}{1-{{y}^{2}}}$ and $Q\left( y \right)=\dfrac{ay}{1-{{y}^{2}}}$

We have Integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}$.

Let $I=\int{P\left( y \right)dy}$

So, we have

$I=\int{\dfrac{y}{1-{{y}^{2}}}dy}$

Put $1-{{y}^{2}}=z$

Differentiating both sides, we get

\[\begin{align}

& -2ydy=dz \\

& \Rightarrow ydy=-\dfrac{dz}{2} \\

\end{align}\]

So, we have

\[\begin{align}

& I=\int{\dfrac{-dz}{2z}} \\

& =-\dfrac{1}{2}\int{\dfrac{dz}{z}} \\

\end{align}\]

We know that $\int{\dfrac{dx}{x}=\ln x+c}$

Using, we get

$I=-\dfrac{1}{2}\ln z$

Returning to the original variable, we get

$\begin{align}

& I=-\dfrac{1}{2}\ln \left( \left| 1-{{y}^{2}} \right| \right) \\

& \Rightarrow I=\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right) \\

\end{align}$

Hence the integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}={{e}^{I}}={{e}^{\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right)}}$

We know that ${{e}^{\ln x}}=x$

Using we get

$IF=\dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}}$.

Hence, options [b] and [d] are correct.

Note: [1]A differential equation when in exact form can be written in the form $du=dv$.

In a Linear differential equation when multiplied by Integrating factor we have $u=y\cdot IF$ and $v=\int{Q\left( x \right)\cdot IFdx}$.

[2] Euler’s criterion for exactness is a direct result of the fact $\dfrac{\partial f}{\partial x\partial y}=\dfrac{\partial f}{\partial y\partial x}$ .

[3] Sometimes, the following identities help in converting a differential equation in the exact form:

[a] $xdy+ydx=d(xy)$

[b] $dx+dy=d(x+y)$

[c] $\dfrac{xdy-ydx}{{{x}^{2}}}=d\left( \dfrac{y}{x} \right)$

[d] $\dfrac{dx}{x}=d\left( \ln x \right)$

[e] $m{{x}^{m-1}}{{y}^{n}}+n{{x}^{m}}{{y}^{n-1}}=d\left( {{x}^{m}}{{y}^{n}} \right)$

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

What is 1 divided by 0 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

Why do noble gases have positive electron gain enthalpy class 11 chemistry CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Write an application to the principal requesting five class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers