1
JEE Main 2024 (Online) 27th January Morning Shift
+4
-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 :
A
$\log _{\frac{2}{3}} 2$
B
$\log _{\frac{4}{3}} 2$
C
$\log _4 3$
D
$\log _3 4$
2
JEE Main 2023 (Online) 15th April Morning Shift
+4
-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 :
A
$\frac{4}{9}$
B
$\frac{32}{9}$
C
$\frac{10}{3}$
D
3
3
JEE Main 2023 (Online) 13th April Morning Shift
+4
-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

A
no point
B
two points
C
infinite number of points
D
one point
4
JEE Main 2023 (Online) 12th April Morning Shift
+4
-1

Let $$y=y(x), y > 0$$, be a solution curve of the differential equation $$\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$$. If $$y(0)=1$$ and $$y(2 \sqrt{2})=\beta$$, then

A
$$e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})$$
B
$$e^{3 \beta^{-1}}=e(5+\sqrt{2})$$
C
$$e^{3 \beta^{-1}}=e(3+2 \sqrt{2})$$
D
$$e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})$$
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