The particular solution of log ( dy / dx )=3 x +4 y , y (0)=0 is

Question

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

Tardigrade Admin · Accepted Answer

( dy / dx )= e 3 x +4 y = e 3 x ⋅ e 4 y ⇒ e -4 y dy = e 3 x dx ⇒ ( e 4 y /-4)=( e 3 x /3)+ c put x =0 We have -(1/4)-(1/3)= c ⇒ c =-(7/12), ∴ ( e -4 y /-4)=( e 3 x /3)-(7/12) ⇒ 7=3 e -4 y +4 e 3 x

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

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

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

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

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

Q. The particular solution of logdxdy​=3x+4y,y(0)=0 is

e3x+3e−4y=4

4e3x−3−4y=3

3e3x+4e4y=7

4e3x+3e−4y=7

dxdy​=e3x+4y=e3x⋅e4y ⇒e−4ydy=e3xdx ⇒−4e4y​=3e3x​+c put x=0 We have −41​−31​=c ⇒c=−127​, ∴−4e−4y​=3e3x​−127​ ⇒7=3e−4y+4e3x

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

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

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

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

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