Solve the differential equation: ${{\left( 1-{{x}^{2}} \right)}^{2}}dy+\left( y\sqrt{1-{{x}^{2}}}-x-\sqrt{1-{{x}^{2}}} \right)dx=0$
Answer
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Hint: To solve the above differential equation we will first try to split x and y terms with dx and dy, that is:
${{\left( 1-{{x}^{2}} \right)}^{2}}dy+y\sqrt{1-{{x}^{2}}}dx-xdx-\sqrt{1-{{x}^{2}}}dx=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x+\left( 1-y \right)\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}-\dfrac{y\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}+\dfrac{y}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}.............\left( 1 \right)$
Now, we can see that the above equation is in the form linear differential equation of first order, that is$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)...............\left( 2 \right)$
And, the solution of above differential equation is given by:
$y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$
Now, after comparing the equation (1) and (2) we will get function P(x) and Q(x) and then we will solve the equation we get to get our required answer.
Complete step-by-step solution:
We will first split the above given differential equation, that is ${{\left( 1-{{x}^{2}} \right)}^{2}}dy+\left( y\sqrt{1-{{x}^{2}}}-x-\sqrt{1-{{x}^{2}}} \right)dx=0$, such that term of x comes with dx and terms of y come with dy.
So, ${{\left( 1-{{x}^{2}} \right)}^{2}}dy+\left( y\sqrt{1-{{x}^{2}}}-x-\sqrt{1-{{x}^{2}}} \right)dx=0$ can be rewritten as:
${{\left( 1-{{x}^{2}} \right)}^{2}}dy+y\sqrt{1-{{x}^{2}}}dx-xdx-\sqrt{1-{{x}^{2}}}dx=0$
$\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{2}}dy=-y\sqrt{1-{{x}^{2}}}dx+xdx+\sqrt{1-{{x}^{2}}}dx$
$\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{2}}dy=\left( x+\left( 1-y \right)\sqrt{1-{{x}^{2}}} \right)dx$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x+\left( 1-y \right)\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}-\dfrac{y\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}+\dfrac{y}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}.............\left( 1 \right)$
Now, we can see that the above equation (1) is in the form of linear differential equation of first order, that is:
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)...............\left( 2 \right)$
So, $P\left( x \right)=\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}},Q\left( x \right)=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}$
And, we know that solution of above equation (2) is given by:
$y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$, here ${{e}^{\int{P\left( x \right)dx}}}$ is also known as integrating factor.
So, integrating factor is:
${{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx}}}............\left( 3 \right)$
Let$x=\cos \theta $, so $dx=-\sin \theta d\theta $
Putting the value of x and dx in equation (3) we will get:
${{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{1}{{{\left( 1-{{\cos }^{2}}\theta \right)}^{\dfrac{3}{2}}}}(-\sin \theta )d\theta }}}$
$\Rightarrow {{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{-\sin \theta }{{{\sin }^{3}}\theta }dx}}}$
$\Rightarrow {{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{-{{\operatorname{cosec}}^{2}}\theta dx}}}$
And we know that $-{{\int{\operatorname{cosec}}}^{2}}\theta d\theta =\cot \theta +c$
So, ${{e}^{\int{P\left( x \right)dx}}}={{e}^{\cot \theta }}$
Now, we will put $x=\cos \theta $, in Q(x):
$Q\left( x \right)=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}$
$Q\left( \theta \right)=\dfrac{\cos \theta }{{{\left( 1-{{\cos }^{2}}\theta \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{\cos }^{2}}\theta \right)}^{\dfrac{3}{2}}}}$
$\Rightarrow Q\left( \theta \right)=\dfrac{\cos \theta }{{{\sin }^{4}}\theta }+\dfrac{1}{{{\sin }^{3}}\theta }$
$\Rightarrow Q\left( \theta \right)=\dfrac{\cos \theta +\sin \theta }{{{\sin }^{4}}\theta }$
Now, we will put the value of ${{e}^{\int{P\left( x \right)dx}}}$, Q(x), and dx in equation $y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$
$\Rightarrow y{{e}^{\cot \theta }}=\int{\left( \dfrac{\cos \theta +\sin \theta }{{{\sin }^{4}}\theta } \right)}\times {{e}^{\cot \theta }}\times \left( -\sin \theta \right)d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{\left( \dfrac{\cos \theta +\sin \theta }{{{\sin }^{3}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{\left( \dfrac{\cos \theta }{{{\sin }^{3}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta -\int{\left( \dfrac{1}{{{\sin }^{2}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{{{\operatorname{cosec}}^{2}}\theta \cot \theta }\times {{e}^{\cot \theta }}d\theta -\int{{{\operatorname{cosec}}^{2}}\theta }\times {{e}^{\cot \theta }}d\theta $
Let $\cot \theta =t$, then $dt=-{{\operatorname{cosec}}^{2}}\theta d\theta $
Now, we will put ‘t’ in place $\cot \theta $ and $dt$ in place of $-{{\operatorname{cosec}}^{2}}\theta d\theta $.
$\therefore y{{e}^{\cot \theta }}=\int{t}\times {{e}^{t}}dt+\int{{{e}^{t}}}dt$
We will use integration by parts as $\int u.v dx = u \int{v} dx - \int(u’)(v)dx$ to solve $\int{t}\times {{e}^{t}}dt$ where $u =t$ and $v= {{e}^{t}}$
$\Rightarrow y{{e}^{\cot \theta }}=t{{e}^{t}}-\int{{{e}^{t}}dt+\int{{{e}^{t}}dt}}$
$\Rightarrow y{{e}^{\cot \theta }}=t{{e}^{t}}+C$
Now, we know that $\cot \theta =t$, so:
$\Rightarrow y{{e}^{\cot \theta }}=\cot \theta \times {{e}^{\cot \theta }}+C$
So, $y=\cot \theta +C{{e}^{-\cot \theta }}$
Now, since we have assumed above that $x=\cos \theta $
So, $\sin \theta =\sqrt{1-{{x}^{2}}}$, so $\cot \theta =\dfrac{x}{\sqrt{1-{{x}^{2}}}}$
So, $y=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+C{{e}^{-\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)}}$
This is our required solution.
Note: Since we have seen that there is a lot of calculation and integrating term in the above solution. So, students are required to be familiar with all integration types and formulas and avoid calculation mistakes and also assume such terms as t which will make our calculation easy.
${{\left( 1-{{x}^{2}} \right)}^{2}}dy+y\sqrt{1-{{x}^{2}}}dx-xdx-\sqrt{1-{{x}^{2}}}dx=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x+\left( 1-y \right)\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}-\dfrac{y\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}+\dfrac{y}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}.............\left( 1 \right)$
Now, we can see that the above equation is in the form linear differential equation of first order, that is$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)...............\left( 2 \right)$
And, the solution of above differential equation is given by:
$y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$
Now, after comparing the equation (1) and (2) we will get function P(x) and Q(x) and then we will solve the equation we get to get our required answer.
Complete step-by-step solution:
We will first split the above given differential equation, that is ${{\left( 1-{{x}^{2}} \right)}^{2}}dy+\left( y\sqrt{1-{{x}^{2}}}-x-\sqrt{1-{{x}^{2}}} \right)dx=0$, such that term of x comes with dx and terms of y come with dy.
So, ${{\left( 1-{{x}^{2}} \right)}^{2}}dy+\left( y\sqrt{1-{{x}^{2}}}-x-\sqrt{1-{{x}^{2}}} \right)dx=0$ can be rewritten as:
${{\left( 1-{{x}^{2}} \right)}^{2}}dy+y\sqrt{1-{{x}^{2}}}dx-xdx-\sqrt{1-{{x}^{2}}}dx=0$
$\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{2}}dy=-y\sqrt{1-{{x}^{2}}}dx+xdx+\sqrt{1-{{x}^{2}}}dx$
$\Rightarrow {{\left( 1-{{x}^{2}} \right)}^{2}}dy=\left( x+\left( 1-y \right)\sqrt{1-{{x}^{2}}} \right)dx$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x+\left( 1-y \right)\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}-\dfrac{y\sqrt{1-{{x}^{2}}}}{{{\left( 1-{{x}^{2}} \right)}^{2}}}$
$\Rightarrow \dfrac{dy}{dx}+\dfrac{y}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}.............\left( 1 \right)$
Now, we can see that the above equation (1) is in the form of linear differential equation of first order, that is:
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)...............\left( 2 \right)$
So, $P\left( x \right)=\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}},Q\left( x \right)=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}$
And, we know that solution of above equation (2) is given by:
$y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$, here ${{e}^{\int{P\left( x \right)dx}}}$ is also known as integrating factor.
So, integrating factor is:
${{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx}}}............\left( 3 \right)$
Let$x=\cos \theta $, so $dx=-\sin \theta d\theta $
Putting the value of x and dx in equation (3) we will get:
${{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{1}{{{\left( 1-{{\cos }^{2}}\theta \right)}^{\dfrac{3}{2}}}}(-\sin \theta )d\theta }}}$
$\Rightarrow {{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{-\sin \theta }{{{\sin }^{3}}\theta }dx}}}$
$\Rightarrow {{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{-{{\operatorname{cosec}}^{2}}\theta dx}}}$
And we know that $-{{\int{\operatorname{cosec}}}^{2}}\theta d\theta =\cot \theta +c$
So, ${{e}^{\int{P\left( x \right)dx}}}={{e}^{\cot \theta }}$
Now, we will put $x=\cos \theta $, in Q(x):
$Q\left( x \right)=\dfrac{x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}$
$Q\left( \theta \right)=\dfrac{\cos \theta }{{{\left( 1-{{\cos }^{2}}\theta \right)}^{2}}}+\dfrac{1}{{{\left( 1-{{\cos }^{2}}\theta \right)}^{\dfrac{3}{2}}}}$
$\Rightarrow Q\left( \theta \right)=\dfrac{\cos \theta }{{{\sin }^{4}}\theta }+\dfrac{1}{{{\sin }^{3}}\theta }$
$\Rightarrow Q\left( \theta \right)=\dfrac{\cos \theta +\sin \theta }{{{\sin }^{4}}\theta }$
Now, we will put the value of ${{e}^{\int{P\left( x \right)dx}}}$, Q(x), and dx in equation $y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}\times {{e}^{\int{P\left( x \right)dx}}}dx$
$\Rightarrow y{{e}^{\cot \theta }}=\int{\left( \dfrac{\cos \theta +\sin \theta }{{{\sin }^{4}}\theta } \right)}\times {{e}^{\cot \theta }}\times \left( -\sin \theta \right)d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{\left( \dfrac{\cos \theta +\sin \theta }{{{\sin }^{3}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{\left( \dfrac{\cos \theta }{{{\sin }^{3}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta -\int{\left( \dfrac{1}{{{\sin }^{2}}\theta } \right)}\times {{e}^{\cot \theta }}d\theta $
$\Rightarrow y{{e}^{\cot \theta }}=-\int{{{\operatorname{cosec}}^{2}}\theta \cot \theta }\times {{e}^{\cot \theta }}d\theta -\int{{{\operatorname{cosec}}^{2}}\theta }\times {{e}^{\cot \theta }}d\theta $
Let $\cot \theta =t$, then $dt=-{{\operatorname{cosec}}^{2}}\theta d\theta $
Now, we will put ‘t’ in place $\cot \theta $ and $dt$ in place of $-{{\operatorname{cosec}}^{2}}\theta d\theta $.
$\therefore y{{e}^{\cot \theta }}=\int{t}\times {{e}^{t}}dt+\int{{{e}^{t}}}dt$
We will use integration by parts as $\int u.v dx = u \int{v} dx - \int(u’)(v)dx$ to solve $\int{t}\times {{e}^{t}}dt$ where $u =t$ and $v= {{e}^{t}}$
$\Rightarrow y{{e}^{\cot \theta }}=t{{e}^{t}}-\int{{{e}^{t}}dt+\int{{{e}^{t}}dt}}$
$\Rightarrow y{{e}^{\cot \theta }}=t{{e}^{t}}+C$
Now, we know that $\cot \theta =t$, so:
$\Rightarrow y{{e}^{\cot \theta }}=\cot \theta \times {{e}^{\cot \theta }}+C$
So, $y=\cot \theta +C{{e}^{-\cot \theta }}$
Now, since we have assumed above that $x=\cos \theta $
So, $\sin \theta =\sqrt{1-{{x}^{2}}}$, so $\cot \theta =\dfrac{x}{\sqrt{1-{{x}^{2}}}}$
So, $y=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+C{{e}^{-\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)}}$
This is our required solution.
Note: Since we have seen that there is a lot of calculation and integrating term in the above solution. So, students are required to be familiar with all integration types and formulas and avoid calculation mistakes and also assume such terms as t which will make our calculation easy.
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