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Question: Solution of $\frac{x+y-1}{x+y-2}dy = \frac{x+y+1}{x+y+2}dx$, given that $y=1$ when $x=1$ is...

Solution of x+y−1x+y−2dy=x+y+1x+y+2dx\frac{x+y-1}{x+y-2}dy = \frac{x+y+1}{x+y+2}dx, given that y=1y=1 when x=1x=1 is

A

ln⁡∣(x−y)2−22∣=2(x+y)\ln \left|\frac{(x-y)^2-2}{2}\right| = 2(x+y)

B

ln⁡∣(x+y)2−22∣=2(x−y)\ln \left|\frac{(x+y)^2-2}{2}\right| = 2(x-y)

C

ln⁡∣(x−y)2+22∣=2(x+y)\ln \left|\frac{(x-y)^2+2}{2}\right| = 2(x+y)

D

ln⁡∣(x+y)2+22∣=2(x−y)\ln \left|\frac{(x+y)^2+2}{2}\right| = 2(x-y)

Answer

ln⁡∣(x+y)2−22∣=2(x−y)\ln \left|\frac{(x+y)^2-2}{2}\right| = 2(x-y)

Explanation

Solution

The given differential equation is: x+y−1x+y−2dy=x+y+1x+y+2dx\frac{x+y-1}{x+y-2}dy = \frac{x+y+1}{x+y+2}dx

Rearrange to find dydx\frac{dy}{dx}: dydx=x+y+1x+y+2⋅x+y−2x+y−1\frac{dy}{dx} = \frac{x+y+1}{x+y+2} \cdot \frac{x+y-2}{x+y-1}

This equation involves the term (x+y)(x+y). Let's use the substitution v=x+yv = x+y. Then, differentiate vv with respect to xx: dvdx=1+dydx  ⟹  dydx=dvdx−1\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1

Substitute vv and dydx\frac{dy}{dx} into the equation: dvdx−1=(v+1)(v−2)(v+2)(v−1)\frac{dv}{dx} - 1 = \frac{(v+1)(v-2)}{(v+2)(v-1)} dvdx=1+v2−v−2v2+v−2\frac{dv}{dx} = 1 + \frac{v^2 - v - 2}{v^2 + v - 2}

Combine the terms on the right side: dvdx=(v2+v−2)+(v2−v−2)v2+v−2\frac{dv}{dx} = \frac{(v^2 + v - 2) + (v^2 - v - 2)}{v^2 + v - 2} dvdx=2v2−4v2+v−2\frac{dv}{dx} = \frac{2v^2 - 4}{v^2 + v - 2}

Separate the variables: v2+v−22(v2−2)dv=dx\frac{v^2 + v - 2}{2(v^2 - 2)} dv = dx

Integrate both sides: ∫v2+v−22(v2−2)dv=∫dx\int \frac{v^2 + v - 2}{2(v^2 - 2)} dv = \int dx 12∫(v2−2)+vv2−2dv=∫dx\frac{1}{2} \int \frac{(v^2 - 2) + v}{v^2 - 2} dv = \int dx 12∫(1+vv2−2)dv=∫dx\frac{1}{2} \int \left(1 + \frac{v}{v^2 - 2}\right) dv = \int dx 12(∫1dv+∫vv2−2dv)=x+C1\frac{1}{2} \left( \int 1 dv + \int \frac{v}{v^2 - 2} dv \right) = x + C_1

For the integral ∫vv2−2dv\int \frac{v}{v^2 - 2} dv, let u=v2−2u = v^2 - 2, so du=2vdv  ⟹  vdv=12dudu = 2v dv \implies v dv = \frac{1}{2}du. ∫vv2−2dv=∫12udu=12ln⁡∣u∣=12ln⁡∣v2−2∣\int \frac{v}{v^2 - 2} dv = \int \frac{1}{2u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|v^2 - 2|

Substitute this back into the main integral: 12(v+12ln⁡∣v2−2∣)=x+C1\frac{1}{2} \left( v + \frac{1}{2} \ln|v^2 - 2| \right) = x + C_1 Multiply by 2: v+12ln⁡∣v2−2∣=2x+2C1v + \frac{1}{2} \ln|v^2 - 2| = 2x + 2C_1 Multiply by 2 again: 2v+ln⁡∣v2−2∣=4x+C2(where C2=4C1)2v + \ln|v^2 - 2| = 4x + C_2 \quad (\text{where } C_2 = 4C_1)

Substitute back v=x+yv = x+y: 2(x+y)+ln⁡∣(x+y)2−2∣=4x+C22(x+y) + \ln|(x+y)^2 - 2| = 4x + C_2

Rearrange the terms to match the options: ln⁡∣(x+y)2−2∣=4x−2(x+y)+C2\ln|(x+y)^2 - 2| = 4x - 2(x+y) + C_2 ln⁡∣(x+y)2−2∣=2x−2y+C2\ln|(x+y)^2 - 2| = 2x - 2y + C_2 ln⁡∣(x+y)2−2∣=2(x−y)+C2\ln|(x+y)^2 - 2| = 2(x-y) + C_2

Now, apply the initial condition y=1y=1 when x=1x=1: ln⁡∣(1+1)2−2∣=2(1−1)+C2\ln|(1+1)^2 - 2| = 2(1-1) + C_2 ln⁡∣22−2∣=0+C2\ln|2^2 - 2| = 0 + C_2 ln⁡∣4−2∣=C2\ln|4 - 2| = C_2 C2=ln⁡2C_2 = \ln 2

Substitute C2=ln⁡2C_2 = \ln 2 back into the general solution: ln⁡∣(x+y)2−2∣=2(x−y)+ln⁡2\ln|(x+y)^2 - 2| = 2(x-y) + \ln 2

Move ln⁡2\ln 2 to the left side: ln⁡∣(x+y)2−2∣−ln⁡2=2(x−y)\ln|(x+y)^2 - 2| - \ln 2 = 2(x-y)

Using the logarithm property ln⁡a−ln⁡b=ln⁡(a/b)\ln a - \ln b = \ln(a/b): ln⁡∣(x+y)2−22∣=2(x−y)\ln \left| \frac{(x+y)^2 - 2}{2} \right| = 2(x-y)