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# How do you solve for the equation $\dfrac{{dy}}{{dx}} = \dfrac{{3{x^2}}}{{{e^{2y}}}}$ that satisfies the initial condition $f\left( 0 \right) = \dfrac{1}{2}$?

Last updated date: 20th Jun 2024
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Hint:To find the given equation is a separable equation in which to solve this differential equation we need to integrate both sides of the equation.

The given differential equation is
$\dfrac{{dy}}{{dx}} = \dfrac{{3{x^2}}}{{{e^{2y}}}}$ ………………………. 1
As the given equation is separable equation, multiply the terms with respect to x and y as
${e^{2y}}dy = 3{x^2}dx$
Hence to solve this differential equation we need to integrate both sides of the equation.
$\int {{e^{2y}}dy} = \int {3{x^2}dx}$
The integration of ${e^{2y}}$ is $\dfrac{1}{2}{e^{2y}}$ and $3{x^2}$is ${x^3} + c$, hence we get
$\dfrac{1}{2}{e^{2y}} = {x^3} + c$…………………… 2
Here we need to find the value of c, hence let us use the given function for $f\left( 0 \right) = \dfrac{1}{2}$in the obtained equation i.e., equation 2 as
$\dfrac{1}{2}{e^{2\left( {\dfrac{1}{2}} \right)}} = {0^3} + c$
Simplifying the equation, the value c is
$c = \dfrac{1}{2}e$
Hence, equation 2 after substituting the value of c is
$\dfrac{1}{2}{e^{2y}} = {x^3} + c$
$\dfrac{1}{2}{e^{2y}} = {x^3} + \dfrac{1}{2}e$
Simplifying the terms, we get
${e^{2y}} = 2{x^3} + e$
$2y = \ln \left( {2{x^3} + e} \right)$
Therefore, the value of $y$ is
$y = \dfrac{1}{2}\ln \left( {2{x^3} + e} \right)$

$\dfrac{{dy}}{{dx}} = f\left( x \right)$
Here $x$ is an independent variable and $y$ is a dependent variable