Question

Solve: \[{x^3}\dfrac{{dy}}{{dx}} = {y^3} + {y^2}\sqrt {{y^2} - {x^2}} \]
1) \[xy = c(x - \sqrt {{y^2} - {x^2}} )\]
2) \[x = c(y + {y^2} - {x^2})\]
3) \[y = c(x + {y^2} - {x^2})\]
4) \[xy = c(y + \sqrt {{y^2} - {x^2}} )\]

Answer 1

Hint: Integration is a reverse process of differentiation, where we reduce the functions into parts. The logarithm is defined as a power to which a number must be raised in order to get some other number. Here, we are given a derivative and we need to find the value of it. We will assume y=vx and using this, we will solve and then we will apply integration on both sides. We will use \[\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = \log (x + \sqrt {{x^2} - {a^2}} ) + c} \] and \[\int {\dfrac{1}{x}dx = \log + c} \] . And then, we will remove the log from both the sides and will get the final output.

Complete step-by-step answer:
We know that integration is also called anti-differentiation.
Given that,
\[{x^3}\dfrac{{dy}}{{dx}} = {y^3} + {y^2}\sqrt {{y^2} - {x^2}} \]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^3} + {y^2}\sqrt {{y^2} - {x^2}} }}{{{x^3}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^3}}}{{{x^3}}} + \dfrac{{{y^2}\sqrt {{y^2} - {x^2}} }}{{{x^3}}}\]
Let \[y = vx\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]
Substitute this value, we will get,
\[ \Rightarrow v + x\dfrac{{dv}}{{dx}} = \dfrac{{{v^3}{x^3}}}{{{x^3}}} + \dfrac{{{v^2}{x^2}\sqrt {{v^2}{x^2} - {x^2}} }}{{{x^3}}}\]
\[ \Rightarrow v + x\dfrac{{dv}}{{dx}} = \dfrac{{{v^3}{x^3}}}{{{x^3}}} + \dfrac{{{v^2}{x^2}\sqrt {{x^2}({v^2} - 1)} }}{{{x^3}}}\]
\[ \Rightarrow v + x\dfrac{{dv}}{{dx}} = \dfrac{{{v^3}{x^3}}}{{{x^3}}} + \dfrac{{{v^2}{x^2} \times x\sqrt {{v^2} - 1} }}{{{x^3}}}\]
\[ \Rightarrow v + x\dfrac{{dv}}{{dx}} = {v^3} + {v^2}\sqrt {{v^2} - 1} \]
By using transposition method, we will move LHS term to RHS, we will get,
\[ \Rightarrow x\dfrac{{dv}}{{dx}} = {v^3} + {v^2}\sqrt {{v^2} - 1} - v\]
\[ \Rightarrow \dfrac{{dv}}{{{v^3} + {v^2}\sqrt {{v^2} - 1} - v}} = \dfrac{{dx}}{x}\]
\[ \Rightarrow \dfrac{{dv}}{{v({v^2} - 1) + {v^2}\sqrt {{v^2} - 1} }} = \dfrac{{dx}}{x}\]
\[ \Rightarrow \dfrac{{dv}}{{v\sqrt {{v^2} - 1} (\sqrt {{v^2} - 1} + v)}} = \dfrac{{dx}}{x}\]
\[ \Rightarrow \dfrac{{dv}}{{v\sqrt {{v^2} - 1} (v + \sqrt {{v^2} - 1} )}} = \dfrac{{dx}}{x}\]
\[ \Rightarrow \dfrac{{v - \sqrt {{v^2} - 1} }}{{v\sqrt {{v^2} - 1} ({v^2} - {v^2} + 1)}}dv = \dfrac{{dx}}{x}\]
\[ \Rightarrow (\dfrac{1}{{\sqrt {{v^2} - 1} }} - \dfrac{1}{v})dv = \dfrac{{dx}}{x}\]
Apply integration on both the sides, we will get,
\[ \Rightarrow \int {\left( {\dfrac{1}{{\sqrt {{v^2} - 1} }} - \dfrac{1}{v}} \right)dv = \int {\dfrac{{dx}}{x}} } \]
\[ \Rightarrow \int {\dfrac{1}{{\sqrt {{v^2} - 1} }}dv - \int {\dfrac{1}{v}} dv = \int {\dfrac{{dx}}{x}} } \]
We know that, \[\int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx = \log (x + \sqrt {{x^2} - {a^2}} ) + c} \] and \[\int {\dfrac{1}{x}dx = \log + logc} \]
Using this, we will get,
\[ \Rightarrow \log (v + \sqrt {{v^2} - 1} ) - \log v = \log x + logc\]
\[ \Rightarrow \log \left( {\dfrac{{v + \sqrt {{v^2} - 1} }}{v}} \right) = \log x + logc\]
We will use this value of \[v = \dfrac{y}{x}\] and substituting this value we will get,
\[ \Rightarrow \log \left( {\dfrac{{\dfrac{y}{x} + \sqrt {{{(\dfrac{y}{x})}^2} - 1} }}{{\dfrac{y}{x}}}} \right) = \log x + logc\]
\[ \Rightarrow \log \left( {\dfrac{{\dfrac{y}{x} + \sqrt {\dfrac{{{y^2} - {x^2}}}{{{x^2}}}} }}{{\dfrac{y}{x}}}} \right) = \log x + c\]
\[ \Rightarrow \log \left( {\dfrac{{\dfrac{y}{x} + \dfrac{1}{x}\sqrt {{y^2} - {x^2}} }}{{\dfrac{y}{x}}}} \right) = \log x + logc\]
\[ \Rightarrow \log \left( {\dfrac{{y + \sqrt {{y^2} - {x^2}} }}{y}} \right) = \log x - logc\]
\[ \Rightarrow \dfrac{{y + \sqrt {{y^2} - {x^2}} }}{y} = \dfrac{x}{c}\]
\[ \Rightarrow y + \sqrt {{y^2} - {x^2}} = \dfrac{x}{c} \times y\]
\[ \Rightarrow c(y + \sqrt {{y^2} - {x^2}} ) = xy\]
Hence, for the given equation, the value is \[xy = c(y + \sqrt {{y^2} - {x^2}} )\] .
So, the correct answer is “Option B”.

Note: Integration is the calculation of an integral. The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. Differentiation which measures the rate of change of any function with respect to its variables. A logarithm is just the opposite function of exponentiation. The difference between log and ln is that log is defined for base 10 and ln is denoted for base e.