The solution of the differential equationdydx-3x-4y-23x-4y-3 is-where C is integration constant

Dydx=(3x 4y 2)(3x 4y 3) ⋯(i) Put 3x 4y=v and so 3 4dydx=dvdx Then, equation (i) becomes: 14(3 dvdx)=v 2v 3 ⇒dvdx=3 4(v 2)v 3 ⇒∫v 3v+1dv= ∫dx ⇒∫(1 4v+1)dv= ∫ ...

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Standard XII

Mathematics

Chain Rule of Differentiation

The solution ...

Question

The solution of the differential equation
$\frac{d y}{d x} = \frac{(3 x - 4 y - 2)}{(3 x - 4 y - 3)}$ is:
(where $C$ is integration constant)

| 3 x - 4 y + 1 | = C e^{x - y}

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| 3 x - 4 y - 2 | = C e^{x + y}

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(3 x - 4 y - 3) = C e^{| 3 x - 4 y |}

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| x + y - 1 | = C e^{3 x - 4 y - 2}

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Solution

The correct option is A $| 3 x - 4 y + 1 | = C e^{x - y}$
$\frac{d y}{d x} = \frac{(3 x - 4 y - 2)}{(3 x - 4 y - 3)} \dots (i)$

Put $3 x - 4 y = v$ and so
$3 - 4 \frac{d y}{d x} = \frac{d v}{d x}$

Then, equation $(i)$ becomes: $\frac{1}{4} (3 - \frac{d v}{d x}) = \frac{v - 2}{v - 3}$
$\Rightarrow \frac{d v}{d x} = 3 - \frac{4 (v - 2)}{v - 3} \Rightarrow \int \frac{v - 3}{v + 1} d v = - \int d x$
$\Rightarrow \int (1 - \frac{4}{v + 1}) d v = - \int d x \Rightarrow v - 4 ln | v + 1 | = - x + 4 c$
$\Rightarrow 4 (x - y - c) = 4 ln | 3 x - 4 y + 1 | \Rightarrow | 3 x - 4 y + 1 | = e^{(x - y - c)}$
$\Rightarrow | 3 x - 4 y + 1 | = C e^{(x - y)}$ where $C = e^{- c} =$ constant.

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Chain Rule of Differentiation

Standard XII Mathematics

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The solution of the differential equation dydx=(3x−4y−2)(3x−4y−3) is: (where C is integration constant)

The solution of the differential equation
$\frac{d y}{d x} = \frac{(3 x - 4 y - 2)}{(3 x - 4 y - 3)}$ is:
(where $C$ is integration constant)