Question

Solve $e^{dy/dx}=x+1$, given that when $x=0,y=3$.

• A. $y=x\ln (x+1)-x+\ln (x+1)+3$
• B. $y=x\ln (x+1)-x+\ln (x-1)+3$
• C. $y=x\ln (x-1)-x+\ln (x+1)+3$
• D. $y=x\ln (x+1)-x^2+\ln (x+1)+3$

Answer 1

The correct option is A

$y=x\ln (x+1)-x+\ln (x+1)+3$

Consider the following integral.

$e^{dy/dx}=x+1$ & x=0,y=3

Taking antilog on both side..

$dy/dx=\ln (x+1)$

$dy=\ln (x+1)dx$

Taking partial integration on both side.

$\int dy=\int ln(x+1)dx....\left(1\right)$

Solve the above eq. Integration by part method

$\int dy=\int 1\ast ln(x+1)dx$

$y=x\ln (x+1)-\int \frac{x}{x+1}dx$

$y=x\ln (x+1)-\int \frac{x+1-1}{x+1}dx$

$=x\ln (x+1)-\int (1-\frac{1}{x+1})dx$

$=x\ln (x+1)-x+\ln (x+1)+C$

Put the value of$x=0$ y=3 we get,

$y=x\ln (x+1)-x+\ln (x+1)+C$

$3=0\ln (0+1)-0+\ln (0+1)+C$

$C=3$

$y=x\ln (x+1)-x+\ln (x+1)+3$

Hence, this is the correct answer.