JEE Main 2024 (Online) 4th April Morning Shift | Differential Equations Question 17 | Mathematics

If the solution $$y=y(x)$$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \mathrm{d} y-(2 x^2+2 x+3) \mathrm{d} x=0$$ satisfies $$y(-1)=-\frac{\pi}{4}$$, then $$y(0)$$ is equal to :

$$-\frac{\pi}{12}$$

$$\frac{\pi}{2}$$

$$\frac{\pi}{4}$$

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Let $\alpha$ be a non-zero real number. Suppose $f: \mathbf{R} \rightarrow \mathbf{R}$ is a differentiable function such that $f(0)=2$ and $\lim\limits_{x \rightarrow-\infty} f(x)=1$. If $f^{\prime}(x)=\alpha f(x)+3$, for all $x \in \mathbf{R}$, then $f\left(-\log _{\mathrm{e}} 2\right)$ is equal to :

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

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

$\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$.

Then, $\left(\frac{1}{\sqrt{2}}+y\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals :

$\frac{4}{4+\sqrt{\mathrm{e}}}$

$\frac{3}{3-\sqrt{\mathrm{e}}}$

$\frac{2}{1+\sqrt{\mathrm{e}}}$

$\frac{1}{2-\sqrt{\mathrm{e}}}$

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\circ} \mathrm{F}$$ and it decreases continuously as per the differential equation $$\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\circ} \mathrm{F}$$, then $$T(45)$$ is equal to

90$$^\circ$$ F

85$$^\circ$$ F

80$$^\circ$$ F

95$$^\circ$$ F

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