1
JEE Main 2024 (Online) 29th January Morning Shift
Numerical
+4
-1

If the solution curve $$y=y(x)$$ of the differential equation $$\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $$\alpha+2 \beta$$ is _________.

2
JEE Main 2024 (Online) 27th January Evening Shift
Numerical
+4
-1

If the solution curve, of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $$5 \beta+\alpha$$ is equal to __________.

3
JEE Main 2024 (Online) 27th January Morning Shift
Numerical
+4
-1
If the solution of the differential equation

$(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3$, is

$\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6$, then $\alpha+2 \beta+3 \gamma$ is equal to ____________.
4
JEE Main 2023 (Online) 13th April Evening Shift
Numerical
+4
-1

If $$y=y(x)$$ is the solution of the differential equation

$$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$$ such that

$$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$$ :