Question

If $y\left( x \right)$ is the differential equation $\dfrac{dy}{dx}+\left( \dfrac{2x+1}{x} \right)y={{e}^{-2x}},x>0$, where $y\left( 1 \right)=\dfrac{1}{2}{{e}^{-2}}$, then:
A. $y\left( x \right)$ is decreasing in $\left( 0,1 \right)$
B. $y\left( x \right)$ is decreasing in $\left( \dfrac{1}{2},1 \right)$
C. $y\left( \ln 2 \right)=\dfrac{\ln 2}{4}$
D. $y\left( \ln 2 \right)=\ln 4$

Answer 1

To solve this question, we will make use of integrating factors. The integrating factor will be chosen in such a way that if we multiply both sides with the integrating factor can make the left-hand side of the general linear differential equation more compact. We will then integrate both sides of this simplified compact linear differential equation to get an equation in x and y and a constant. We will then apply the given condition $y\left( 1 \right)=\dfrac{1}{2}{{e}^{-2}}$ to find the constant. Once we get the final equation, we can verify which of the given options are correct.

Complete step by step answer:
A linear differential equation of first order given in form as shown below can be solved by finding the integrating factor of the differential equation. The general linear differential equation is given below
$\dfrac{dy}{dx}+f(x)y=g(x)$
The equation of this form can be solved by finding integrating factor and following procedures given below:
First we find the integrating factor$(I)$ with the formula given below. This integrating factor will be a function of x since $f(x)$ and $g(x)$ also are function of x
$I={{e}^{\int{f(x)dx}}}$
Multiplying both sides of general first order linear differential equation with this integrating factor we get,
${{e}^{\int f\left( x \right)dx}}\left( \dfrac{dy}{dx}+f\left( x \right)y \right)={{e}^{\int f\left( x \right)dx}}g\left( x \right)$
${{e}^{\int f\left( x \right)dx}}\dfrac{dy}{dx}+f\left( x \right)y{{e}^{\int f\left( x \right)dx}}={{e}^{\int f\left( x \right)dx}}g\left( x \right)$
Left hand side can be written compactly as shown below with important formula $\dfrac{d}{dx}\left( {{e}^{\int f\left( x \right)dx}} \right)=f\left( x \right){{e}^{\int f\left( x \right)dx}}$ we can write it as simply as possible:
$\dfrac{d}{dx}\left( {{e}^{\int f\left( x \right)dx}} \right)=f\left( x \right){{e}^{\int f\left( x \right)dx}}$
On integrating both sides with respect to x we get the complete solution of the above differential equation:-
$y{{e}^{\int f\left( x \right)dx}}=\int {{e}^{\int f\left( x \right)dx}}g\left( x \right)dx+c$, where c is arbitrary constant
Applying the same process with given question
$\dfrac{dy}{dx}+\left( \dfrac{2x+1}{x} \right)y={{e}^{-2x}}$
We find the Integrating factor(I) with the method stated above
$I={{e}^{\int{\dfrac{2x+1}{x}dx}}}={{e}^{\int{\left( 2+\dfrac{1}{x} \right)dx}}}={{e}^{2x+\int{\dfrac{dx}{x}}}}={{e}^{2x+\ln (x)}}={{e}^{2x}}{{e}^{\ln (x)}}=x{{e}^{2x}}$
Using Integrating factor to solve differential equation,
$yx{{e}^{2x}}=\int{x{{e}^{2x}}{{e}^{-2x}}dx}+c$
$yx{{e}^{2x}}=\int{x{{e}^{2x-2x}}dx}+c$
$yx{{e}^{2x}}=\int{xdx}+c$
And this is the solution we were solving for,
$yx{{e}^{2x}}=\dfrac{{{x}^{2}}}{2}+c$
Given $y\left( 1 \right)=\dfrac{1}{2}{{e}^{-2}}$
To find value of constant use given values mentioned in the question,
We can substitute the given boundary conditions to find the arbitrary constant,At $x=1$we get$y=\dfrac{1}{2}{{e}^{-2}}$
$y{{e}^{2}}=\dfrac{{{1}^{2}}}{2}+c$
$\dfrac{1}{2}{{e}^{-2}}^{+2}=\dfrac{1}{2}+c$
We get $c=0$
We have obtained the particular solution for the differential equation mentioned above after substituting value of constant which is –
$yx{{e}^{2x}}=\dfrac{{{x}^{2}}}{2}$
$y{{e}^{2x}}=\dfrac{x}{2}$
We can see the particular solution for the given differential equation:-
$y=\dfrac{x}{2}{{e}^{-2x}}$
In order to determine whether function y(x) is a decreasing function if so in what interval. We will differentiate the given function wrt to x and if ${y}'(x)$ is greater than 0 then it is an increasing function else a decreasing function.
We know the formula $\dfrac{d({{e}^{x}})}{dx}={{e}^{x}}$ and $\dfrac{d\left[ f(x)g(x) \right]}{dx}=f'(x)g(x)+f(x)g'(x)$
Using above formula we get the derivative as:
${y}'=\dfrac{{{e}^{-2x}}}{2}+\dfrac{x{{e}^{-2x}}}{2}(-2)$
${y}'=\dfrac{{{e}^{-2x}}}{2}-x{{e}^{-2x}}$
Given below is derivative of the function y:
${y}'=\left( \dfrac{1}{2}-x \right){{e}^{-2x}}$
If ${y}'<0$ then the function is a decreasing function.
$\left( \dfrac{1}{2}-x \right){{e}^{-2x}}<0$
Hence, the given function is decreasing for values of x given below,
$x>\dfrac{1}{2}$

So, the correct answer is “Option B”.

Note: In case above general linear differential equation is of form $\dfrac{dx}{dy}+f(y)x=g(y)$ then in that case integrating factor becomes $I={{e}^{\int{f(y)dy}}}$ and general solution comes in form $x{{e}^{\int f\left( y \right)dy}}=\int {{e}^{\int f\left( y \right)dy}}g\left( y \right)dy+c$.