Question

The differential equation $\left(x{y}^3\right(1+\cos x)-y)dx+xdy=0$ represents the curve $\frac{x^2}{2y^2}=\frac{x^3}{b}+x^2\sin x+cx\cos x-d\sin x+k$ then $b+c+d$ is

• A. $6$
• B. $7$
• C. $9$
• D. $10$

Answer 1

The correct option is B $7$

$\frac{dy}{dx}+y^3(1+\cos x)-\frac{y}{x}=0$
$\frac{1}{y^3}\frac{dy}{dx}-\frac{1}{y^2}\cdot \frac{1}{x}=-(1+\cos x)$
Let $-\frac{1}{y^2}=u\Rightarrow \frac{2}{y^3}\frac{dy}{dx}=\frac{du}{dx}$
$\frac{du}{dx}+\frac{2}{x}\cdot u=-2(1+\cos x)$
I.F. $=e^{\int \frac{2}{x}dx}=x^2$
$u\cdot x^2=\int -2(1+\cos x)x^{2}dx$
$-\frac{x^2}{y^2}=-\frac{2x^3}{3}-2x^2\sin x-4x\cos x+4\sin x+k$
$\frac{x^2}{2y^2}=\frac{x^3}{3}+x^2\sin x+2x\cos x-2\sin x+k$

$\therefore b=3,c=2,d=2$
Thus, $b+c+d=3+2+2=7$