Let f : R \to R be a differentiable function with f(0) = 0. If y = f(x) satisfies the differential equation dy \over dx = (2 + 5y)(5y - 2) , then the value of \mathop \lim \limits_n \to - \infty f(x) is ...........

Differential Equations PYQ — JEE Advanced 2018 · Shift 2 · Q28 with Solution

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Correct answer: 0.4

We are given the differential equation $\frac{dy}{dx}=(2+5y)(5y-2).$ Since $(2+5y)(5y-2)=25y^2-4,$ this is an autonomous differential equation: $\frac{dy}{dx}=25y^2-4.$
We are also given the initial condition $f(0)=0.$ So the solution curve passes through $(0,0)$ .
We need the limiting value of $f(x)$ as the independent variable tends to $-\infty$ . (There appears to be a typo in the question statement writing $n\to -\infty$ ; it should be $x\to -\infty$ .)
First find the equilibrium solutions by setting $\frac{dy}{dx}=0.$ Thus, $25y^2-4=0\implies y=\pm \frac{2}{5}.$ So the constant solutions are $y=\frac25,\qquad y=-\frac25.$
Analyze the sign of $\dfrac{dy}{dx}$ :

If $-\frac25<y<\frac25$ , then $25y^2-4<0$ , so $\dfrac{dy}{dx}<0$ .
If $y>\frac25$ or $y<-\frac25$ , then $25y^2-4>0$ .

Since the solution starts at $y(0)=0$ , which lies in $\left(-\frac25,\frac25\right)$ , we have $\frac{dy}{dx}<0$ near $x=0$ . Hence as $x$ increases, $y$ decreases; equivalently, as $x$ decreases toward $-\infty$ , the solution moves upward.

So for $x<0$ , the solution increases and is bounded above by the equilibrium value $\frac25$ . This strongly suggests $\lim_{x\to -\infty} y(x)=\frac25.$

To confirm, solve explicitly by separation of variables: $\frac{dy}{25y^2-4}=dx.$ Factor the denominator: $25y^2-4=(5y-2)(5y+2).$ Using partial fractions, $\frac{1}{25y^2-4}=\frac{1}{4}\left(\frac{1}{5y-2}-\frac{1}{5y+2}\right).$ So $\int \frac{dy}{25y^2-4}=\int dx.$ Thus, $\frac14\int\left(\frac{1}{5y-2}-\frac{1}{5y+2}\right)dy=x+C.$

Since

\qquad \int \frac{dy}{5y+2}=\frac15\ln|5y+2|,$$ we get $$\frac{1}{20}\ln\left|\frac{5y-2}{5y+2}\right|=x+C.$$ Therefore, $$\ln\left|\frac{5y-2}{5y+2}\right|=20x+C_1,$$ so $$\frac{5y-2}{5y+2}=Ce^{20x}.$$ 7. Use the initial condition $y(0)=0$: $$\frac{5(0)-2}{5(0)+2}=C\cdot e^0 \implies \frac{-2}{2}=C \implies C=-1.$$ Hence, $$\frac{5y-2}{5y+2}=-e^{20x}.$$ Now solve for $y$: $$5y-2=-e^{20x}(5y+2),$$ $$5y+5ye^{20x}=2-2e^{20x},$$ $$5y(1+e^{20x})=2(1-e^{20x}),$$ $$y=\frac{2}{5}\cdot \frac{1-e^{20x}}{1+e^{20x}}.$$ 8. Now take the limit as $x\to -\infty$: $$e^{20x}\to 0.$$ Therefore, $$\lim_{x\to -\infty} y(x)=\frac{2}{5}\cdot \frac{1-0}{1+0}=\frac25=0.4.$$ 9. Hence, $$\boxed{\lim_{x\to -\infty} f(x)=\frac25=0.4}.$$

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View written solutionFree

Correct answer: 0.4

We are given the differential equation $\frac{dy}{dx}=(2+5y)(5y-2).$ Since $(2+5y)(5y-2)=25y^2-4,$ this is an autonomous differential equation: $\frac{dy}{dx}=25y^2-4.$
We are also given the initial condition $f(0)=0.$ So the solution curve passes through $(0,0)$ .
We need the limiting value of $f(x)$ as the independent variable tends to $-\infty$ . (There appears to be a typo in the question statement writing $n\to -\infty$ ; it should be $x\to -\infty$ .)
First find the equilibrium solutions by setting $\frac{dy}{dx}=0.$ Thus, $25y^2-4=0\implies y=\pm \frac{2}{5}.$ So the constant solutions are $y=\frac25,\qquad y=-\frac25.$
Analyze the sign of $\dfrac{dy}{dx}$ :

If $-\frac25<y<\frac25$ , then $25y^2-4<0$ , so $\dfrac{dy}{dx}<0$ .
If $y>\frac25$ or $y<-\frac25$ , then $25y^2-4>0$ .

So for $x<0$ , the solution increases and is bounded above by the equilibrium value $\frac25$ . This strongly suggests $\lim_{x\to -\infty} y(x)=\frac25.$

To confirm, solve explicitly by separation of variables: $\frac{dy}{25y^2-4}=dx.$ Factor the denominator: $25y^2-4=(5y-2)(5y+2).$ Using partial fractions, $\frac{1}{25y^2-4}=\frac{1}{4}\left(\frac{1}{5y-2}-\frac{1}{5y+2}\right).$ So $\int \frac{dy}{25y^2-4}=\int dx.$ Thus, $\frac14\int\left(\frac{1}{5y-2}-\frac{1}{5y+2}\right)dy=x+C.$

Since

\qquad \int \frac{dy}{5y+2}=\frac15\ln|5y+2|,$$ we get $$\frac{1}{20}\ln\left|\frac{5y-2}{5y+2}\right|=x+C.$$ Therefore, $$\ln\left|\frac{5y-2}{5y+2}\right|=20x+C_1,$$ so $$\frac{5y-2}{5y+2}=Ce^{20x}.$$ 7. Use the initial condition $y(0)=0$: $$\frac{5(0)-2}{5(0)+2}=C\cdot e^0 \implies \frac{-2}{2}=C \implies C=-1.$$ Hence, $$\frac{5y-2}{5y+2}=-e^{20x}.$$ Now solve for $y$: $$5y-2=-e^{20x}(5y+2),$$ $$5y+5ye^{20x}=2-2e^{20x},$$ $$5y(1+e^{20x})=2(1-e^{20x}),$$ $$y=\frac{2}{5}\cdot \frac{1-e^{20x}}{1+e^{20x}}.$$ 8. Now take the limit as $x\to -\infty$: $$e^{20x}\to 0.$$ Therefore, $$\lim_{x\to -\infty} y(x)=\frac{2}{5}\cdot \frac{1-0}{1+0}=\frac25=0.4.$$ 9. Hence, $$\boxed{\lim_{x\to -\infty} f(x)=\frac25=0.4}.$$

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