Question

Find the general solution of the differential equation\[\dfrac{{dy}}{{dx}} = 2xy\] from the following option.
A. \[y = c{e^{{x^2}}}\]
B. \[{y^2} = 2{x^2} + c\]
C. \[y = {e^{ - {x^2}}} + c\]
D. \[y = {x^2} + c\]

Answer 1

Hint: The given differential equation is a linear first order differential equation. There are two variables x and y. To find a general solution of the differential equation use the variable separation method. Separate the terms containing x and y. Then integrate the equation.

Formula Used: \[\begin{array}{l}\int {\dfrac{1}{x}dx} = \log x + c\\\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\end{array}\]
Note that, the constant c is an arbitrary constant and n cannot be zero.
The law of exponential \[{e^{m + n}} = {e^m}{e^n}\]

Complete step by step solution: The given differential equation is \[\dfrac{{dy}}{{dx}} = 2xy\].
First separate x containing terms and y containing terms and simplify the equation. To separate variables by multiplying \[dx\]on both sides.
\[\begin{array}{l}\dfrac{{dy}}{{dx}} \cdot dx = 2xy \cdot dx\\dy = 2xydx\end{array}\]
Now, divide by y on both sides of the equation.
\[\dfrac{{dy}}{y} = 2xdx\]
As variables are separated now. So integrate both sides of the equation.
\[\begin{array}{l}\int {\dfrac{{dy}}{y}} = \int {2xdx} \\\log y + {c_1} = 2\int {xdx} \\\log y + {c_1} = 2 \cdot \dfrac{{{x^2}}}{2} + {c_2}\end{array}\]
 Simplify the equation as follows.
\[\begin{array}{l}\log y + {c_1} = {x^2} + {c_2}\\\log y = {x^2} + {c_2} - {c_1}\end{array}\]
Here the constant is arbitrary. So, we can write \[{c_2} - {c_1} = c\].
\[\log y = {x^2} + c\]
Isolate y by applying exponential function on both sides of the equation.
\[y = {e^{{x^2} + c}}\]
Apply exponential rules and simplify.
\[y = {e^{{x^2}}} \cdot {e^c}\]
 The constant is arbitrary so we can say as \[{e^c} = c\]. Now simplify the equation as follows.
\[y = c{e^{{x^2}}}\]
Hence the general solution for the differential equation is \[y = c{e^{{x^2}}}\].

Option ‘A’ is correct

Note: The common mistake happens while solving this type of question is integration of \[\dfrac{1}{y}\]is written as \[\dfrac{1}{{{y^2}}}\], which is wrong. Also, one more mistake can happen here is while isolating y. \[y = \log ({x^2} + c)\] this is the wrong method.