JEE Main 2023 (Online) 15th April Morning Shift | Differential Equations Question 38 | Mathematics

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

JEE Main 2023 (Online) 12th April Morning Shift

MCQ (Single Correct Answer)

-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

$$e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})$$

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

$$e^{3 \beta^{-1}}=e(3+2 \sqrt{2})$$

$$e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})$$

JEE Main 2023 (Online) 11th April Evening Shift

MCQ (Single Correct Answer)

-1

Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :

$$\frac{693}{128}$$

$$\frac{697}{128}$$

$$\frac{637}{128}$$

$$\frac{679}{128}$$

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