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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Differential Equations

BITSAT 2020 · Mathematics · Differential Equations

The solutions of (x+y+1) d y=d x are

The solutions of (x+y+1)dy=dx(x+y+1) d y=d x are

Options

  1. Ax+y+2=Ceyx+y+2=C e^{y}
  2. Bx+y+4=Clog⁡yx+y+4=C \log y
  3. Clog⁡(x+y+2)=Cy\quad \log (x+y+2)=C y
  4. Dlog⁡(x+y+2)=C−y\log (x+y+2)=C-y

Correct answer

A. x+y+2=Ceyx+y+2=C e^{y}

Step-by-step solution

Putting x+y+1=u,x+y+1=u, we have du=dx+dyd u=d x+d y and the given equation reduces to u(du−dx)=dx⇒uduu+1=dx⇒u−log⁡(u+1)=x⇒log⁡(x+y+2)=y+u(d u-d x)=d x\Rightarrow \quad \frac{u d u}{u+1}=d x \quad \Rightarrow \quad u-\log (u+1)=x\Rightarrow \log (x+y+2)=y+ constant ⇒x+y+2=Cey\Rightarrow x+y+2=C e^{y}
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