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Question: If \(y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } \) , then show that \(\...

If y=x+y+x+y+.........∞y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } , then show that dydx=y2−x2y3−2xy−1\dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{2{y^3} - 2xy - 1}}.

Explanation

Solution

In order to solve the equation, check that where the term yy has repeated and, then mark them as yy , we will get a shorter equation. Then to remove the square root, square both the sides. Then differentiate the terms with respect to xx and separate the variables on one side.

Formula used:
1. dyndx=nyn−1dydx\dfrac{{d{y^n}}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}
2. dxndx=nxn−1\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}
3. d(uv)dx=ud(v)dx+vd(u)dx\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + v\dfrac{{d\left( u \right)}}{{dx}}

Complete step by step solution:
We are given the equation y=x+y+x+y+.........∞y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } .
We can see that after the second root, from the third root yy is getting repeated, as y=x+y+x+y+.........∞y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } .
So, writing x+y+.........∞\sqrt {x + \sqrt {y + .........\infty } } as yy , we get the equation as:
y=x+y+x+y+.........∞ ⇒y=x+y+y  y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } \\\ \Rightarrow y = \sqrt {x + \sqrt {y + y} } \\\
It can be written as:
y=x+y+y ⇒y=x+2y  y = \sqrt {x + \sqrt {y + y} } \\\ \Rightarrow y = \sqrt {x + \sqrt {2y} } \\\
Squaring both the sides, in order to remove the square root, as we know that (x)2=x{\left( {\sqrt x } \right)^2} = x.
So,
y=x+2y ⇒(y)2=(x+2y)2 ⇒y2=x+2y  y = \sqrt {x + \sqrt {2y} } \\\ \Rightarrow {\left( y \right)^2} = {\left( {\sqrt {x + \sqrt {2y} } } \right)^2} \\\ \Rightarrow {y^2} = x + \sqrt {2y} \\\
Subtracting both the sides of the above equation by xx:
y2=x+2y ⇒y2−x=x+2y−x ⇒y2−x=2y  {y^2} = x + \sqrt {2y} \\\ \Rightarrow {y^2} - x = x + \sqrt {2y} - x \\\ \Rightarrow {y^2} - x = \sqrt {2y} \\\

Since, we have one more square root on the right side, so squaring both the sides, and we get:
y2−x=2y ⇒(y2−x)2=(2y)2 ⇒(y2−x)2=2y  {y^2} - x = \sqrt {2y} \\\ \Rightarrow {\left( {{y^2} - x} \right)^2} = {\left( {\sqrt {2y} } \right)^2} \\\ \Rightarrow {\left( {{y^2} - x} \right)^2} = 2y \\\
Opening the brackets using (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab.
(y2−x)2=2y =>y4+x2−2xy2=2y  {\left( {{y^2} - x} \right)^2} = 2y \\\ = > {y^4} + {x^2} - 2x{y^2} = 2y \\\
Subtracting both the sides by 2y2y , we get:
y4+x2−2xy2=2y ⇒y4+x2−2xy2−2y=2y−2y ⇒y4+x2−2xy2−2y=0  {y^4} + {x^2} - 2x{y^2} = 2y \\\ \Rightarrow {y^4} + {x^2} - 2x{y^2} - 2y = 2y - 2y \\\ \Rightarrow {y^4} + {x^2} - 2x{y^2} - 2y = 0 \\\
Since, we obtained a sorted, so differentiating both the sides of the above equation with respect to xx :

y4+x2−2xy2−2y=0 ⇒d(y4+x2−2xy2−2y)dx=d0dx ⇒d(y4+x2−2xy2−2y)dx=0  {y^4} + {x^2} - 2x{y^2} - 2y = 0 \\\ \Rightarrow \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = \dfrac{{d0}}{{dx}} \\\ \Rightarrow \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = 0 \\\

Splitting the terms for differentiation:

d(y4+x2−2xy2−2y)dx=0 ⇒dy4dx+dx2dx−d(2xy2)dx−d(2y)dx=0  \dfrac{{d\left( {{y^4} + {x^2} - 2x{y^2} - 2y} \right)}}{{dx}} = 0 \\\ \Rightarrow \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - \dfrac{{d\left( {2x{y^2}} \right)}}{{dx}} - \dfrac{{d\left( {2y} \right)}}{{dx}} = 0 \\\

Taking out the constants from the parenthesis:

\dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - \dfrac{{d\left( {2x{y^2}} \right)}}{{dx}} - \dfrac{{d\left( {2y} \right)}}{{dx}} = 0 \\\ \Rightarrow \dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - 2\dfrac{{d\left( {x{y^2}} \right)}}{{dx}} - 2\dfrac{{d\left( y \right)}}{{dx}} = 0 \\\ $$ …….(A) We would solve each operand separately: Starting with $$\dfrac{{d{y^4}}}{{dx}}$$. Since, we know that $$\dfrac{{d{y^n}}}{{dx}} = n{y^{n - 1}}\dfrac{{dy}}{{dx}}$$. So, from this we get: $$\dfrac{{d{y^4}}}{{dx}} = 4{y^{4 - 1}}\dfrac{{dy}}{{dx}} = 4{y^3}\dfrac{{dy}}{{dx}}$$. ………………..(1) Then we have $$\dfrac{{d{x^2}}}{{dx}}$$ : Since, we know that $$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$$. So, from this we get: $$\dfrac{{d{x^2}}}{{dx}} = 2{x^{2 - 1}} = 2x$$. ………………..(2) Then we have $$\dfrac{{d\left( {x{y^2}} \right)}}{{dx}}$$ : Since, we know that it is in product form, and it can differentiated as $$\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{d\left( v \right)}}{{dx}} + v\dfrac{{d\left( u \right)}}{{dx}}$$. Comparing $$\left( {uv} \right)$$ with $$\left( {x{y^2}} \right)$$, we get $$u = x,v = {y^2}$$ So, differentiating it from the product rule, we get:

\dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = x\dfrac{{d\left( {{y^2}} \right)}}{{dx}} + {y^2}\dfrac{{d\left( x \right)}}{{dx}} \\
\Rightarrow \dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = x.2y\dfrac{{dy}}{{dx}} + {y^2}\left( 1 \right) \\
\Rightarrow \dfrac{{d\left( {x{y^2}} \right)}}{{dx}} = 2xy\dfrac{{dy}}{{dx}} + {y^2} \\

Then we have $$\dfrac{{dy}}{{dx}}$$ : Since, we can see that it cannot be further simplified. Therefore, it is $$\dfrac{{dy}}{{dx}}$$ only ………………..(4) Substituting (1), (2), (3) and (4) in A, we get:

\dfrac{{d{y^4}}}{{dx}} + \dfrac{{d{x^2}}}{{dx}} - 2\dfrac{{d\left( {x{y^2}} \right)}}{{dx}} - 2\dfrac{{d\left( y \right)}}{{dx}} = 0 \\
\Rightarrow 4{y^3}\dfrac{{dy}}{{dx}} + 2x - 2\left( {2xy\dfrac{{dy}}{{dx}} + {y^2}} \right) - 2\dfrac{{dy}}{{dx}} = 0 \\

Openingtheparenthesisabove,andweget:Opening the parenthesis above, and we get:

4{y^3}\dfrac{{dy}}{{dx}} + 2x - 2\left( {2xy\dfrac{{dy}}{{dx}} + {y^2}} \right) - 2\dfrac{{dy}}{{dx}} = 0 \\
\Rightarrow 4{y^3}\dfrac{{dy}}{{dx}} + 2x - 4xy\dfrac{{dy}}{{dx}} - 2{y^2} - 2\dfrac{{dy}}{{dx}} = 0 \\

Taking $$2$$ common: $$2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right) = 0$$ Dividing both the sides by $$2$$:

2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right) = 0 \\
\Rightarrow \dfrac{{2\left( {2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}}} \right)}}{2} = 0 \\
\Rightarrow 2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}} = 0 \\

Taking coefficient of $$\dfrac{{dy}}{{dx}}$$ in one parenthesis:

2{y^3}\dfrac{{dy}}{{dx}} + x - 2xy\dfrac{{dy}}{{dx}} - {y^2} - \dfrac{{dy}}{{dx}} = 0 \\
\Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} = 0 \\

Adding both the sides by $${y^2}$$ and subtracting $$x$$ , we get:

\left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} = 0 \\
\Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} + x - {y^2} + {y^2} - x = {y^2} - x \\
\Rightarrow \left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}} = {y^2} - x \\

Dividing both the sides by $$\left( {2{y^3} - 2xy - 1} \right)$$ , and we get:

\dfrac{{\left( {2{y^3} - 2xy - 1} \right)\dfrac{{dy}}{{dx}}}}{{\left( {2{y^3} - 2xy - 1} \right)}} = \dfrac{{{y^2} - x}}{{\left( {2{y^3} - 2xy - 1} \right)}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{\left( {2{y^3} - 2xy - 1} \right)}} \\

Which is proved. **Therefore, if $y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + .........\infty } } } } $ , then $\dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - x}}{{2{y^3} - 2xy - 1}}$.** **Hence, proved.** **Note:** 1\. Always preferred to go step by step for ease, otherwise there is a huge chance of mistakes to arise. 2\. It’s important to square the terms to remove the roots, otherwise, it would become much more complicated to differentiate.