Question

Let $f:R\to R$ be a differentiable function $f\left( 0 \right)=0$. If $y=f\left( x \right)$ satisfies the differential equation $\dfrac{dy}{dx}=\left( 2+5y \right)\left( 5y-2 \right),$ then the value of $\displaystyle \lim_{x \to -\infty}f\left( x \right)=k$. Find the value of $5k.$

Answer 1

Use separation of variable to initiate the integration. Use a partial fraction method to convert the integrand to the standard form of indefinite integration formula of $\dfrac{1}{x}$. Then use the given initial condition to find out the value of constant occurred during integration. Finally use the concept limiting approach to determine the value of $k.$

Complete step-by-step answer:

The given differential equation is $\dfrac{dy}{dx}=\left( 2+5y \right)\left( 5y-2 \right)$. Let us separate the variables at different sides, $\begin{aligned} & \dfrac{dy}{dx}=\left( 2+5y \right)\left( 5y-2 \right) \\\ & \Rightarrow \dfrac{dy}{\left( 5y+2 \right)\left( 5y-2 \right)}=dx....(1) \\\ \end{aligned}$
We use the partial fraction method to integrate the above. Let us suppose
\begin{aligned} & I=\dfrac{1}{\left( 5y+2 \right)\left( 5y-2 \right)}=\dfrac{A}{5y+2}+\dfrac{B}{5y-2} \\\ & \Rightarrow 1=A\left( 5y-2 \right)+B\left( 5y+2 \right)....(2) \\\ \end{aligned}$$$$$ where A$and$B$are real numbers. We assign$y=\dfrac{2}{5}$in equation (2) to get$B=\dfrac{1}{4}$and then assign$y=-\dfrac{2}{5}$to get$A=\dfrac{-1}{4}. So we get $$$$ I=\dfrac{\dfrac{-1}{4}}{5y+2}+\dfrac{\dfrac{1}{4}}{5y-2}=\dfrac{1}{4}\left( \dfrac{1}{5y-2}-\dfrac{1}{5y+2} \right)$$$$$

Substituting above in equation (1) we proceed to integrate using substitution of variable and
the formula $\int{\dfrac{1}{x}}=\ln |x|+c$ ,

& \dfrac{dy}{\left( 5y+2 \right)\left( 5y-2 \right)}=dx \\\ & \Rightarrow \dfrac{1}{4}\left( \dfrac{1}{5y-2}-\dfrac{1}{5y+2} \right)dy=dx \\\ & \Rightarrow \dfrac{1}{4}\left( \int{\dfrac{dy}{5y-2}}-\int{\dfrac{dy}{5y+2}} \right)=\int{dx} \\\ & \Rightarrow \dfrac{1}{4}\left( \dfrac{\ln \left| 5y-2 \right|}{5}-\dfrac{\ln \left| 5y+2 \right|}{5} \right)=x+c \\\ & \Rightarrow \dfrac{1}{20}\ln \left| \dfrac{5y-2}{5y+2} \right|=x+c....(3) \\\ \end{aligned}$$ where $c$ is a real constant. We use the given initial condition to find out the value of $c$. The given initial condition is $f\left( 0 \right)=0$ which means $x=0\Rightarrow y=0$. Putting it in the above equation (3) we get, $$\dfrac{1}{20}\ln \left| \dfrac{5\cdot 0-2}{5\cdot 0+2} \right|=0+c\Rightarrow c=0$$ Putting the value of $c$ in the equation (3) we get, $$\dfrac{1}{20}\ln \left| \dfrac{5y-2}{5y+2} \right|=x+0\Rightarrow 20x=\ln \left| \dfrac{5y-2}{5y+2} \right|\Rightarrow {{e}^{20}}^{x}=\left| \dfrac{5y-2}{5y+2} \right|$$ We take opening the modulus bracket as with same sign as exponential is always positive, $${{e}^{20}}^{x}=\dfrac{5y-2}{5y+2}\Rightarrow y=\dfrac{2\left( {{e}^{20x}}+1 \right)}{5\left( 1-{{e}^{20x}} \right)}\Rightarrow f\left( x \right)=\dfrac{2{{e}^{20x}}\left( +1 \right)}{5\left( 1-{{e}^{20x}} \right)}$$ Taking both side the required limit $x\to -\infty $, $$\displaystyle \lim_{x \to -\infty}f\left( x \right)=\displaystyle \lim_{x \to -\infty}\dfrac{2{{e}^{20x}}\left( +1 \right)}{5\left( 1-{{e}^{20x}} \right)}=\dfrac{2}{5}=0.4$$ Alternatively, If $x\to -\infty $ then $$20x=\ln \left| \dfrac{5y-2}{5y+2} \right|\to -\infty $$ . Now, $$\begin{aligned} & \ln \left| \dfrac{5y-2}{5y+2} \right|\to -\infty \\\ & \Rightarrow \left| \dfrac{5y-2}{5y+2} \right|\to {{e}^{-\infty }}=0 \\\ & \Rightarrow y\to \dfrac{2}{5} \\\ & \Rightarrow \displaystyle \lim_{x \to -\infty}f\left( x \right)\to \dfrac{2}{5}=0.4 \\\ \end{aligned}$$ As given in the question, $\displaystyle \lim_{x \to -\infty}f\left( x \right)\to k $. Comparing with the above result $k=0.4$. As asked in the question we need to find the value of $5k$. So the value of $5k$ is $5k=5\times 0.4=2$. **Note:** The question tests your knowledge of the solution of differential equations with initial condition. It also requires the knowledge of limiting value. Careful substitution and intuitive use of limiting value helps to arrive at the correct result.