Question

If the solution of the differential equation $\frac{dy}{dx}=\frac{(1+y{)}^2}{x(1+y)-x^2}$ is $\frac{\lambda +y}{x}+c=\ln (1+y),$ then the value of $(10\lambda +6)$ is equal to

• A. 10
• B. 16
• C. 12
• D. 14

Answer 1

The correct option is B 16

From given differential equations

$\frac{dx}{dy}=\frac{x(1+y)-x^2}{(1+y{)}^2}$

$\Rightarrow \frac{dx}{dy}-\frac{1}{(1+y)}x=-\frac{x^2}{(1+y{)}^2}$

$\Rightarrow -\frac{1}{x^2}\frac{dx}{dy}+\frac{1}{x}.\frac{1}{(1+y)}=\frac{1}{(1+y{)}^2},\text{Put}\frac{1}{x}=z$

$\Rightarrow -\frac{1}{x^2}\frac{dx}{dy}=\frac{dz}{dy}$

$\therefore \frac{dz}{dy}+\frac{1}{(1+y)}z=+\frac{1}{(1+y{)}^2}$ ...(i)

I.f. $=e^{\int \frac{dx}{1+y}}=(1+y)$

$\therefore$~~Solutions of equations (i) is

$z\times (1+y)=\int \frac{+dy}{(1+y)}$

$\Rightarrow \frac{1+y}{x}+c=\ln (1+y)$

From passing, we get $\lambda =1$

$\therefore 10\lambda +6=10\times 1+6=16$