Question

solve dy/dx=X+y+1/x-y​

Answer 1

Answer:

tan−1x−yx+y+1+ln{(y+x+1)2+(x−y)2−−−−−−−−−−−−−−−−−−√}=C′

Step-by-step explanation:

dyx+y+1=dxx−y=dy+λdx(x+y+1)+λ(x−y)=d(y+λx)(1−λ)y+(1+λ)x+1=d((1−λ)y+λ(1−λ)x)(1−λ){(1−λ)y+(1+λ)x+1}  

For consistency in numerator and denominator,

λ(1−λ)=1+λ⟹λ=±i

Both values of \lambda satisfy the equation . Hence

d((1−λ1)y+(1+λ1)x+1)(1−λ1){(1−λ1)y+(1+λ1)x+1}=d((1−λ2)y+(1+λ2)x+1)(1−λ2){(1−λ2)y+(1+λ2)x+1}⟹

(1−λ2)d((1−λ1)y+(1+λ1)x+1){(1−λ1)y+(1+λ1)x+1}=(1−λ1)d((1−λ2)y+(1+λ2)x+1){(1−λ2)y+(1+λ2)x+1}

(1+i)d((y+x+1)+i(x−y)){y+x+1)+i(x−y)}=((1−i)d((y+x+1)−i(x−y)){y+x+1)−i(x−y)}

Integrating

(1+i)ln|(y+x+1)+i(x−y)|−(1−i)ln|(y+x+1)−i(x−y)|=C

ln|(y+x+1)+i(x−y)|−ln|(y+x+1)−i(x−y)|+i{ln|(y+x+1)+i(x−y)|+ln|(y+x+1)−i(x−y)|}

Designating (y+x+1)+i(x−y)=Reitwhere

R=(y+x+1)2+(x−y)2−−−−−−−−−−−−−−−−−−√and t=tan−1x−yx+y+1

lnR+it−(lnR−it)+ilnR2=C

t+lnR=C′

tan−1x−yx+y+1+ln{(y+x+1)2+(x−y)2−−−−−−−−−−−−−−−−−−√}=C′