Q.25. Let be the solution of the differential equation , . If , then the greatest integer less than

Q: What is the correct answer for JEE Main Mathematics Q25?

The correct answer is 3. Sol. The given differential equation is $(x^2 - x\sqrt{x^2-1})\frac{dy}{dx} + y(x - \sqrt{x^2-1}) - x = 0$. Divide by $(x^2 - x\sqrt{x^2-1})$: $\frac{dy}{dx} + \frac{y(x - \sqrt{x^2-1})}{x(x - \sqrt{x^2-1})} - \frac{x}{x(x - \sqrt{x^2-1})} = 0$. This simplifies to $\frac{dy}{dx} + \frac{y}{x} - \frac{1}{x - \sqrt{x^2-1}} = 0$. So, $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2-1}}$. This is

Q: Q.25. Let be the solution of the differential equation , . If , then the greatest integer less than is ________.

The correct answer is 3. Sol. The given differential equation is $(x^2 - x\sqrt{x^2-1})\frac{dy}{dx} + y(x - \sqrt{x^2-1}) - x = 0$. Divide by $(x^2 - x\sqrt{x^2-1})$: $\frac{dy}{dx} + \frac{y(x - \sqrt{x^2-1})}{x(x - \sqrt{x^2-1})} - \frac{x}{x(x - \sqrt{x^2-1})} = 0$. This simplifies to $\frac{dy}{dx} + \frac{y}{x} - \frac{1}{x - \sqrt{x^2-1}} = 0$. So, $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2-1}}$. This is

Question

Q.25. Let  be the solution of the differential equation , . If , then the greatest integer less than  is ________.

FirstInTest · Accepted Answer

Sol. The given differential equation is $(x^2 - x\sqrt{x^2-1})\frac{dy}{dx} + y(x - \sqrt{x^2-1}) - x = 0$. Divide by $(x^2 - x\sqrt{x^2-1})$: $\frac{dy}{dx} + \frac{y(x - \sqrt{x^2-1})}{x(x - \sqrt{x^2-1})} - \frac{x}{x(x - \sqrt{x^2-1})} = 0$. This simplifies to $\frac{dy}{dx} + \frac{y}{x} - \frac{1}{x - \sqrt{x^2-1}} = 0$. So, $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2-1}}$. This is a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x - \sqrt{x^2-1}}$. The integrating factor (I.F.) is $e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is $y \cdot 	ext{I.F.} = \int Q(x) \cdot 	ext{I.F.} dx$. So, $y \cdot x = \int \frac{1}{x - \sqrt{x^2-1}} \cdot x dx$. To evaluate the integral $\int \frac{x}{x - \sqrt{x^2-1}} dx$, multiply numerator and denominator by $x + \sqrt{x^2-1}$: $\int \frac{x(x + \sqrt{x^2-1})}{x^2 - (x^2-1)} dx = \int x(x + \sqrt{x^2-1}) dx = \int (x^2 + x\sqrt{x^2-1}) dx$. This integral can be split: $\int x^2 dx + \int x\sqrt{x^2-1} dx$. $\int x^2 dx = \frac{x^3}{3}$. For $\int x\sqrt{x^2-1} dx$, let $u = x^2-1$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. The integral becomes $\int \frac{1}{2} \sqrt{u} du = \frac{1}{2} \frac{u^{3/2}}{3/2} = \frac{1}{3} u^{3/2} = \frac{1}{3} (x^2-1)^{3/2}$. So, $yx = \frac{x^3}{3} + \frac{1}{3} (x^2-1)^{3/2} + C$. Given $y(1) = 1$. Substitute $x=1, y=1$: $1 \cdot 1 = \frac{1^3}{3} + \frac{1}{3} (1^2-1)^{3/2} + C \Rightarrow 1 = \frac{1}{3} + 0 + C \Rightarrow C = 1 - \frac{1}{3} = \frac{2}{3}$. So the solution is $yx = \frac{x^3}{3} + \frac{1}{3} (x^2-1)^{3/2} + \frac{2}{3}$. We need to find $y(\sqrt{5})$. Substitute $x = \sqrt{5}$: $y(\sqrt{5}) \cdot \sqrt{5} = \frac{(\sqrt{5})^3}{3} + \frac{1}{3} ((\sqrt{5})^2-1)^{3/2} + \frac{2}{3}$. $y(\sqrt{5}) \cdot \sqrt{5} = \frac{5\sqrt{5}}{3} + \frac{1}{3} (5-1)^{3/2} + \frac{2}{3} = \frac{5\sqrt{5}}{3} + \frac{1}{3} (4)^{3/2} + \frac{2}{3} = \frac{5\sqrt{5}}{3} + \frac{1}{3} (8) + \frac{2}{3} = \frac{5\sqrt{5} + 8 + 2}{3} = \frac{5\sqrt{5} + 10}{3}$. So, $y(\sqrt{5}) = \frac{5\sqrt{5} + 10}{3\sqrt{5}} = \frac{5 + 10/\sqrt{5}}{3} = \frac{5 + 2\sqrt{5}}{3}$. Now, approximate the value: $\sqrt{5} \approx 2.236$. $y(\sqrt{5}) \approx \frac{5 + 2(2.236)}{3} = \frac{5 + 4.472}{3} = \frac{9.472}{3} \approx 3.157$. The greatest integer less than $y(\sqrt{5})$ is 3.

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions