A particular solution of log ((d y/d x))=3 x+4 y, y(0)=0 is

Question

A particular solution of log ((d y/d x))=3 x+4 y, y(0)=0 is (A) e3 x+3 e-4 y=4 (B) 4 e3 x-3 e-4 y=3 (C) 3 e3 x+4 e-4 y=7 (D) 4 e3 x+3 e-4 y=7

Tardigrade Admin · Accepted Answer

Given, log ((d y/d x))=3 x+4 y ⇒ (d y/d x)=e3 x e4 y ⇒ e-4 y d y=e3 x d x On integrating both sides, we get -4 e-4 y=3 e3 x+c At x=0, y=0, -4=3+c ⇒ c=-7 ∴ Solution is 4 e-4 y+3 e3 x=7

Q. A particular solution of $lo g (\frac{d y}{d x}) = 3 x + 4 y, y (0) = 0$ is

$e^{3 x} + 3 e^{- 4 y} = 4$

$4 e^{3 x} - 3 e^{- 4 y} = 3$

$3 e^{3 x} + 4 e^{- 4 y} = 7$

$4 e^{3 x} + 3 e^{- 4 y} = 7$

Q. A particular solution of log(dxdy​)=3x+4y,y(0)=0 is

e3x+3e−4y=4

4e3x−3e−4y=3

3e3x+4e−4y=7

4e3x+3e−4y=7

Given, log(dxdy​)=3x+4y ⇒dxdy​=e3xe4y ⇒e−4ydy=e3xdx On integrating both sides, we get −4e−4y=3e3x+c At x=0,y=0, −4=3+c⇒c=−7 ∴ Solution is 4e−4y+3e3x=7

Q. A particular solution of $lo g (\frac{d y}{d x}) = 3 x + 4 y, y (0) = 0$ is

$e^{3 x} + 3 e^{- 4 y} = 4$

$4 e^{3 x} - 3 e^{- 4 y} = 3$

$3 e^{3 x} + 4 e^{- 4 y} = 7$

$4 e^{3 x} + 3 e^{- 4 y} = 7$