
Select the general solution of the differential equation \[ydx - xdy = {x^2}ydx\]from the following options.
A \[y{e^{{x^2}}} = c{x^2}\]
B \[y{e^{ - {x^2}}} = c{x^2}\]
C \[{y^2}{e^{{x^2}}} = c{x^2}\]
D \[{y^2}{e^{ - {x^2}}} = c{x^2}\]
Answer
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Hint: The general solution of the differential equation can be found by using the variable separation method. Simplify the equation to separate the x containing terms and y containing terms and integrate it to find a general solution.
Formula Used: Integrating formula
\[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\end{array}\]
Here, c is an arbitrary constant and n can not be zero.
Complete step by step solution: The given differential equation is \[ydx - xdy = {x^2}ydx\].
Simplify the equation by subtracting \[ydx\]on both sides of the equation.
\[ - xdy = {x^2}ydx - ydx\]
Now, Multiply the equation by -1 on both sides of the equation.
\[xdy = ydx - {x^2}ydx\]
Take \[ydx\]common from the right side of the equation.
\[xdy = ydx\left( {1 - {x^2}} \right)\]
Now divide by x on both sides of the equation.
\[dy = \dfrac{y}{x}dx\left( {1 - {x^2}} \right)\]
Divide both sides of the equation by y to separate the x containing terms and y containing terms.
\[\dfrac{{dy}}{y} = \left( {\dfrac{{1 - {x^2}}}{x}} \right)dx\]
Integrate both sides of the equation.
\[\begin{array}{l}\int {\dfrac{{dy}}{y}} = \int {\dfrac{{1 - {x^2}}}{x}dx} \\\log y = \int {\dfrac{1}{x}dx} - \int {xdx} \\\log y = \log x - \dfrac{{{x^2}}}{2} + \log c\end{array}\]
Multiply both sides of the equation by \[2\]
\[2\log y = 2\log x - {x^2} + 2\log c\]
Now add \[{x^2}\]on both sides of the equation.
\[2\log y + {x^2} = 2\log x + 2\log c\]
Subtract \[2\log y\]on both sides of the equation.
\[\begin{array}{l}{x^2} = 2\log x + 2\log c - 2\log y\\{x^2} = \log {x^2} + \log c - \log {y^2}\\{x^2} = \log \dfrac{{c{x^2}}}{{{y^2}}}\end{array}\]
Apply exponential function on both sides of the equation.
\[{e^{{x^2}}} = \dfrac{{c{x^2}}}{{{y^2}}}\]
Multiply both sides of the equation by \[{y^2}\]
\[{y^2}{e^{{x^2}}} = c{x^2}\]
The solution of the equation is \[{y^2}{e^{{x^2}}} = c{x^2}\].
Option ‘C’ is correct
Note: The common mistake in this type of question is first taking direct integration of both sides without separating variables. This leads to the wrong solution.
Formula Used: Integrating formula
\[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\end{array}\]
Here, c is an arbitrary constant and n can not be zero.
Complete step by step solution: The given differential equation is \[ydx - xdy = {x^2}ydx\].
Simplify the equation by subtracting \[ydx\]on both sides of the equation.
\[ - xdy = {x^2}ydx - ydx\]
Now, Multiply the equation by -1 on both sides of the equation.
\[xdy = ydx - {x^2}ydx\]
Take \[ydx\]common from the right side of the equation.
\[xdy = ydx\left( {1 - {x^2}} \right)\]
Now divide by x on both sides of the equation.
\[dy = \dfrac{y}{x}dx\left( {1 - {x^2}} \right)\]
Divide both sides of the equation by y to separate the x containing terms and y containing terms.
\[\dfrac{{dy}}{y} = \left( {\dfrac{{1 - {x^2}}}{x}} \right)dx\]
Integrate both sides of the equation.
\[\begin{array}{l}\int {\dfrac{{dy}}{y}} = \int {\dfrac{{1 - {x^2}}}{x}dx} \\\log y = \int {\dfrac{1}{x}dx} - \int {xdx} \\\log y = \log x - \dfrac{{{x^2}}}{2} + \log c\end{array}\]
Multiply both sides of the equation by \[2\]
\[2\log y = 2\log x - {x^2} + 2\log c\]
Now add \[{x^2}\]on both sides of the equation.
\[2\log y + {x^2} = 2\log x + 2\log c\]
Subtract \[2\log y\]on both sides of the equation.
\[\begin{array}{l}{x^2} = 2\log x + 2\log c - 2\log y\\{x^2} = \log {x^2} + \log c - \log {y^2}\\{x^2} = \log \dfrac{{c{x^2}}}{{{y^2}}}\end{array}\]
Apply exponential function on both sides of the equation.
\[{e^{{x^2}}} = \dfrac{{c{x^2}}}{{{y^2}}}\]
Multiply both sides of the equation by \[{y^2}\]
\[{y^2}{e^{{x^2}}} = c{x^2}\]
The solution of the equation is \[{y^2}{e^{{x^2}}} = c{x^2}\].
Option ‘C’ is correct
Note: The common mistake in this type of question is first taking direct integration of both sides without separating variables. This leads to the wrong solution.
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