Question

What is the solution of the differential equation \[ {x^2}dy = - 2xydx\]?
A. \[x{y^2} = c\]
B. \[{x^2}{y^2} = c\]
C. \[{x^2}y = c\]
D. \[xy = c\]

Answer 1

Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral formulas. In the end, apply the properties of a logarithm to get the solution of the differential equation.

Formula used:
\[\int {\dfrac{1}{x}dx = \log x} \]
The logarithmic properties:
\[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\]
\[n\log\left( a \right) = \log\left( {{a^n}} \right)\]
If \[\log\left( a \right) = \log\left( b \right)\], then \[a = b\].

Complete step by step solution:
The given differential equation is \[ {x^2}dy = - 2xydx\].

Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[\dfrac{{ dy}}{y} = \dfrac{{ - 2x}}{{{x^2}}}dx\]
\[ \Rightarrow \dfrac{{ dy}}{y} = \dfrac{{ - 2}}{x}dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{ dy}}{y}} = \int {\dfrac{{ - 2}}{x}dx} \]
Take the constant value out from the right-hand side.
\[\int {\dfrac{{ dy}}{y}} = - 2\int {\dfrac{1}{x}dx} \]
Now solve the integrals by using the integration formula \[\int {\dfrac{1}{x}dx = log x} \].
\[\log y = - 2\log x + \log c\]
Simplify the above equation.
\[\log y + 2\log x = \log c\]
\[ \Rightarrow \log y + \log {x^2} = \log c\]
Apply the logarithmic property of addition \[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\].
We get,
\[log {x^2}y = \log c\]
Equating both sides, we get
\[{x^2}y = c\]
Hence the correct option is C.

Note: Students often confused with formula of sum of two logarithmic functions. Sometimes they used an incorrect formula \[\log\left( a \right) + \log\left( b \right) = \log\left( {a+b} \right)\]. But the correct formula is \[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\].