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Question

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[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

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] Eulerâ€™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] Eulerâ€™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] $m{{x}^{m-1}}{{y}^{n}}+n{{x}^{m}}{{y}^{n-1}}=d\left( {{x}^{m}}{{y}^{n}} \right)$

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