TG EAPCET 2025 (Online) 2nd May Morning Shift | Differential Equations Question 11 | Mathematics

If $\cos x \frac{d y}{d x}=y \sin x-1, x \neq(2 n+1) \frac{\pi}{2}, n \in Z$ is the differential equation corresponding to the curve $y=f(x)$ and $f(0)=1$, then $f(x)$

$(1-x) \sec x$

$(1-x) \cos x$

$x+\cos x$

$x+\sec x$

TG EAPCET 2025 (Online) 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $2 d x+d y=(6 x y+4 x-3 y) d x$ is

$2 \log |2 x-1|=3 y^2+4 y+C$

$\log |3 y+2|=3 x^2-3 x+C$

$\log |3 y+2|=x^2-x+C$

$\log |2 x-1|=3 y^2-4 y+C$

TG EAPCET 2024 (Online) 11th May Morning Shift

MCQ (Single Correct Answer)

-0

If the equation of the curve which passes through the point $(1,1)$ satisfies the differential equation $\frac{d y}{d x}=\frac{2 x-5 y+3}{5 x+2 y-3}$, then the equation of that curve is

$x^{2}+5 x y-y^{2}+3 x-3 y-5=0$

$x^{2}+5 x y-y^{2}+3 x+3 y-11=0$

$x^{2}-5 x y-y^{2}-3 x-3 y+11=0$

$x^{2}-5 x y-y^{2}+3 x+3 y-1=0$

TG EAPCET 2024 (Online) 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\left(6 x^{2}-2 x y-18 x+3 y\right) d x-\left(x^{2}-3 x\right) d y=0$ is

$2 x^{3}-x^{2} y-9 x^{2}+3 x y+C=0$

$4 x^{3}-2 x^{2} y-6 x^{2}+6 x y+C=0$

$2 x^{2}-4 x y-y^{2}-x+3 y+C=0$

$3 x^{2}+5 x y-2 y^{2}-4 x-2 y+C=0$

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