The general solution of d y/d x=x+y+1/x+y-1 is

Correct Answer is : y=x+log⁡(x+y)+cy=x+ (x+y)+cy=x+log(x+y)+c, where c is a constant of integration.

y=x+ (x+y)+c, where c is a constant of integration.

y=x- (x+y)+c, where c is a constant of integration.

y=x- (2 x+y)+c, where c is a constant of integration.

y=x²+ (x+y)+c, where c is a constant of integration.

Correct Answer is : y=x+log⁡(x+y)+cy=x+ (x+y)+cy=x+log(x+y)+c, where c is a constant of integration.

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MHTCET 10 May 2024 Shift 1 Paper

Question : 141 of 150

Marks: +1, -0

The general solution of

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

Solution:

Concept:

Use the substitution

v = x+y

to reduce the given equation to a separable form.

Explanation:

Let

v = x+y

Then

\frac{dv}{dx} = 1 + \frac{dy}{dx}

, so

\frac{dy}{dx} = \frac{dv}{dx} - 1

Substituting into the given equation:

\frac{dv}{dx} - 1 = \frac{v+1}{v-1}

\Rightarrow \frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{2v}{v-1}

Separate the variables:

\frac{v-1}{2v} \, dv = dx

Integrate both sides:

\int \frac{v-1}{2v} \, dv = \int dx

\frac{1}{2} \int \left(1 - \frac{1}{v}\right) dv = x + C_1

\frac{1}{2} (v - \ln v) = x + C_1

v - \ln v = 2x + 2C_1

Substitute back

v = x+y

x+y - \ln(x+y) = 2x + 2C_1

y = x + \ln(x+y) + c, \text{ where } c = 2C_1

Answer:

Option A:

y=x+\log (x+y)+c

, where

c

is a constant of integration.

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