Question

$y-x\frac{dy}{dx}=b(1+x^2\frac{dy}{dx}).$
Solve the above differential equation.

• A. $y=k(y-b)(1+bx).$
• B. $y=k(y+b)(1-bx).$
• C. $y=k(y+b)(1+bx).$
• D. $y=k(y-b)(1-bx).$

Answer 1

The correct option is A $y=k(y-b)(1+bx).$

After rearranging the terms, we get
$y-b=\frac{dy}{dx}(b{x}^2+x)$
$\frac{dx}{x(bx+1)}=\frac{dy}{y-b}$
$\int \frac{bx+1-bx}{x(bx+1)}.dx=ln(y-b)$
$\int \frac{1}{x}-\frac{b}{bx+1}.dx=ln(y-b)$
$ln\left(x\right)-ln(bx+1)+ln\left(c\right)=ln(y-b)$
$\frac{cx}{bx+1}=y-b$