6. ⇒ (MHT CET 2021 21th September Morning Shift )

The general solution of the differential equation $\frac{d y}{d x} = \frac{x + 2 y - 1}{x + 2 y + 1}$ is

A. $3 (x + y) + 4 \log | 3 x + 6 y - 1 | = K$

B. $3 (x - y) + 4 \log | 3 x + 6 y - 1 | = K$

C. $6 (- x + y) + 4 \log | 3 x + 6 y - 1 | = K$

D. $6 (x + y) + 4 \log | 3 x + 6 y - 1 | = K$

Correct Option is (c)

$\frac{d y}{d x} = \frac{x + 2 y - 1}{x + 2 y + 1}$

Put $x + 2 y = t \Rightarrow 1 + 2 \frac{d y}{d x} = \frac{d t}{d x} \Rightarrow \frac{d y}{d x} = \frac{(\frac{d t}{d x} - 1)}{2}$

$\begin{aligned} \frac{(\frac{d t}{d x} - 1)}{2} = \frac{t - 1}{t + 1} \Rightarrow \frac{d t}{d x} - 1 = \frac{2 t - 2}{t + 1} \\ ∴ \frac{d t}{d x} = \frac{2 t - 2}{t + 1} + 1 = \frac{3 t - 1}{t + 1} \\ ∴ \frac{t + 1}{3 t - 1} d t = \int d x \\ ∴ \frac{1}{3} \int \frac{3 (t + 1)}{3 t - 1} d t = \int d x \Rightarrow \frac{1}{3} \int \frac{3 t - 1 + 4}{3 t - 1} d t = \int d x \\ ∴ \frac{1}{3} \int d t + \frac{4}{3} \int \frac{d t}{3 t - 1} = \int d x \\ ∴ \frac{t}{3} + \frac{4}{3} \frac{\log | 3 t - 1 |}{3} = x + c_{1} \\ ∴ \frac{x + 2 y}{3} + \frac{4}{3} \frac{\log | 3 (x + 2 y) - 1 |}{3} = x + c_{1} \\ ∴ 3 x + 6 y + 4 \log | 3 x + 6 y - 1 | = 9 x + 9 c_{1} ∴ 6 (- x + y) + 4 \log | 3 x + 6 y - 1 | = K \end{aligned}$

7. ⇒ (MHT CET 2021 20th September Evening Shift )

The general solution of $(x \frac{d y}{d x} - y) \sin \frac{y}{x} = x^{3} e^{x}$ is

A. $e^{x} (x - 1) + \cos \frac{y}{x} + c = 0$

B. $x e^{x} + \cos \frac{y}{x} + c = 0$

C. $e^{x} (x + 1) + \cos \frac{y}{x} + c = 0$

D. $e x^{x} - \cos \frac{y}{x} + c = 0$

Correct Option is (A)

$\begin{aligned} (x \frac{d y}{d x} - y) \sin \frac{y}{x} = x^{3} e^{x} . . . . . (i) \\ Put \frac{y}{x} = t \Rightarrow \frac{x \frac{d y}{d x} - y}{x^{2}} = \frac{d t}{d x} \end{aligned}$

Hence eq. (i) becomes

$\begin{aligned} x^{2} (\frac{dt}{dx}) \sin t = x^{3} e^{x} \Rightarrow (\frac{dt}{dx}) \sin t = {xe}^{x} \\ ∴ \int \sin t dt = \int {xe}^{x} dx \\ ∴ - \cos t = {xe}^{x} - \int e^{x} dx = {xe}^{x} - e^{x} + c \\ ∴ - \cos (\frac{y}{x}) = e^{x} (x - 1) + c \end{aligned}$

8. ⇒ (MHT CET 2021 20th September Morning Shift )

The general solution of the differential equation $\frac{d y}{d x} = \frac{x + y + 1}{x + y - 1}$ is given by

A. $y = x \log (x + y) + c$

B. $x - y = \log (x + y) + c$

C. $x + y = \log (x + y) + c$

D. $y = x + \log (x + y) + c$

Correct Option is (D)

$\begin{aligned} \frac{d y}{d x} = \frac{x + y + 1}{x + y - 1} \\ Put x + y = u \Rightarrow 1 + \frac{d y}{d x} = \frac{d u}{d x} \\ ∴ \frac{d u}{d x} - 1 = \frac{u + 1}{u - 1} \Rightarrow \frac{d u}{d x} = \frac{u + 1}{u - 1} + 1 = \frac{2 u}{u - 1} \\ ∴ (\frac{u - 1}{u}) d u = 2 d x \\ Integrating both sides, we get \\ \int d u - \int \frac{d u}{u} = \int 2 d x \\ ∴ u - \log u = 2 x + c \\ ∴ x + y - \log (x + y) = 2 x + c \Rightarrow y = x + \log (x + y) + c \end{aligned}$

9. ⇒ (MHT CET 2021 20th September Morning Shift )

The general solution of the differential equation $x + y \frac{d y}{d x} = \sec (x^{2} + y^{2})$ is

A. $\sin (x^{2} + y^{2}) = 2 x + c$

B. $\sin (x^{2} + y^{2}) + 2 x = c$

C. $\sin (x^{2} + y^{2}) + x = c$

D. $\cos (x^{2} + y^{2}) = 2 x + c$

Correct Option is (A)

We have, $x + y \frac{d y}{d x} = \sec (x^{2} + y^{2})$

Put $x^{2} + y^{2} = u \Rightarrow 2 x + 2 y \frac{d y}{d x} = \frac{d u}{d x}$

$\begin{aligned} ∴ x + y \frac{d y}{d x} = \frac{1}{2} \frac{d u}{d x} \\ ∴ \frac{1}{2} \frac{d u}{d x} = \sec u \\ ∴ \int \frac{d u}{\sec u} = \int 2 d x \\ ∴ \sin u = 2 x + c \Rightarrow \sin (x^{2} + y^{2}) = 2 x + c \end{aligned}$

⇒Defferential Equation