JEE Main 2024 (Online) 27th January Evening Shift | Differential Equations Question 34 | Mathematics

If $$y=y(x)$$ is the solution curve of the differential equation $$\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x>2, y(4)=\frac{3}{2}$$ and the slope of the curve is never zero, then the value of $$y(10)$$ equals :

$$\frac{3}{1+(8)^{1 / 4}}$$

$$\frac{3}{1-(8)^{1 / 4}}$$

$$\frac{3}{1-2 \sqrt{2}}$$

$$\frac{3}{1+2 \sqrt{2}}$$

JEE Main 2024 (Online) 27th January Morning Shift

MCQ (Single Correct Answer)

-1

Let $x=x(\mathrm{t})$ and $y=y(\mathrm{t})$ be solutions of the differential equations $\frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{a} x=0$ and $\frac{\mathrm{d} y}{\mathrm{dt}}+\mathrm{by}=0$ respectively, $\mathrm{a}, \mathrm{b} \in \mathbf{R}$. Given that $x(0)=2 ; y(0)=1$ and $3 y(1)=2 x(1)$, the value of $\mathrm{t}$, for which $x(\mathrm{t})=y(\mathrm{t})$, is :

$\log _{\frac{2}{3}} 2$

$\log _{\frac{4}{3}} 2$

$\log _4 3$

$\log _3 4$

JEE Main 2023 (Online) 15th April Morning Shift

MCQ (Single Correct Answer)

-1

Let $x=x(y)$ be the solution of the differential equation

$2(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y>-1$

with $x\left(e^{4}-2\right)=1$. Then $x\left(e^{9}-2\right)$ is equal to :

$\frac{4}{9}$

$\frac{32}{9}$

$\frac{10}{3}$

JEE Main 2023 (Online) 13th April Morning Shift

MCQ (Single Correct Answer)

-1

Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be the solution curves of the differential equation $$\frac{d y}{d x}=y+7$$ with initial conditions $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then the curves $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ intersect at

no point

two points

infinite number of points

one point

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