Question

What is a solution to the differential equation $\dfrac{dy}{dx}=1-0.2y$?

Answer 1

For the equation in this form, we can use a variable separable method. With the help of this invariable separable method, we can find the solution of the given differential equation in a simple and fast manner. After separating all $x$terms to one side and all $y$ to the other side we will integrate on both sides.

Complete step by step solution:
From the question we were given that
\Rightarrow $$$$\dfrac{dy}{dx}=1-0.2y……………(1)
This equation can be written as $\dfrac{dy}{dx}=1-\dfrac{2}{10}y$
After simplification we get $\dfrac{dy}{dx}=1-\dfrac{1}{5}y$
So now let us take LCM on RHS. LCM of 1,5 is 5
So, we get equation as $\dfrac{dy}{dx}=\dfrac{5-y}{5}$……………(2)
Now using variable separable method lets us bring all $x$terms to one side and all $y$ to other side.
Now we will get the equation as
$\dfrac{dy}{5-y}=\dfrac{dx}{5}$……………..(3)
Multiply the whole equation with $-1$, we get
$\dfrac{dy}{5-y}\left( -1 \right)=\dfrac{dx}{5}\left( -1 \right)$
On simplification we get
\Rightarrow $$$$\dfrac{dy}{y-5}=-\dfrac{dx}{5}……………..(4)
So now let us do integration on sides, now we get
\Rightarrow $$$$\int{\dfrac{dy}{y-5}=\int{-}\dfrac{dx}{5}} …………….(5)
From the basic integration formula, we know that $\int{\dfrac{1}{x}dx}=\ln x+c$ and $\int{a}dx=a\times x+{{c}_{1}}$ where $a,c,{{c}_{1}}$ are constants.
So, we can write $\int{\dfrac{dy}{y-5}}=\ln \left( y-5 \right)+c$ and $\int{-}\dfrac{dx}{5}=-\dfrac{1}{5}\times x+{{c}_{1}}$ now put these values in equation (5)
So, we will get the equation as
\Rightarrow $$$$\ln \left( y-5 \right)+c=-\dfrac{1}{5}\times x+{{c}_{1}}
Add $\dfrac{1}{5}\times x$ on both sides, so we get the equation as
\Rightarrow $$$$\ln \left( y-5 \right)+\dfrac{1}{5}\times x+c=-\dfrac{1}{5}\times x+{{c}_{1}}+\dfrac{1}{5}\times x
After simplification we get the equation as $\ln \left( y-5 \right)+\dfrac{x}{5}=c$ ( where $c$is the common constant for both $c$and ${{c}_{1}}$).
So we got $\ln \left( y-5 \right)+\dfrac{x}{5}=c$………………..(6)
Now let us rise both the LHS and RHS to the power $e$
So, we get equation as
\Rightarrow $$$${{e}^{\ln \left( y-5 \right)+\dfrac{x}{5}}}={{e}^{c}}……………..(7)
From the basic algebraic formula we know that the ${{x}^{a+b}}={{x}^{a}}\times {{x}^{b}}$ we can write ${{e}^{\ln \left( y-5 \right)+\dfrac{x}{5}}}={{e}^{\ln \left( y-5 \right)}}\times {{e}^{\dfrac{x}{5}}}$
So put ${{e}^{\ln \left( y-5 \right)+\dfrac{x}{5}}}={{e}^{\ln \left( y-5 \right)}}\times {{e}^{\dfrac{x}{5}}}$ in equation(7)
So, we get new equation as
$\Rightarrow$ ${{e}^{\ln \left( y-5 \right)}}\times {{e}^{\dfrac{x}{5}}}={{e}^{c}}$
\Rightarrow $$$${{e}^{\ln \left( y-5 \right)}}\times {{e}^{\dfrac{x}{5}}}=c [${{e}^{c}}$ is also a constant]
We know from the basic logarithmic functions that ${{e}^{\ln x}}=x$ we can write ${{e}^{\ln \left( y-5 \right)}}=y-5$
Now let us multiply with ${{e}^{\dfrac{-x}{5}}}$ on both sides.
Now we will get the equation as
\Rightarrow $$$$y-5=c\times {{e}^{\dfrac{-x}{5}}}
Add 5 on both sides, we get equation as
\Rightarrow $$$$y-5+5=c\times {{e}^{\dfrac{-x}{5}}}+5
On simplification we get
\Rightarrow $$$$y=c\times {{e}^{\dfrac{-x}{5}}}+5
So $y=c\times {{e}^{\dfrac{-x}{5}}}+5$ is the solution of the given equation $\dfrac{dy}{dx}=1-0.2y$

Note: Students should do calculations correct to avoid errors in the final answer. Students should use proper formula.In case of misconception it will lead to a huge mistake while finding the solution of differential equation $\dfrac{dy}{dx}=1-0.2y$.