Question

If $y=f(x)$ is a solution of $\frac{dy}{dx}=(y-kx)^{2}+K$ when $x=0$ and $y=1$, then $f(2)=$

• A. $2K+1$
• B. $2K-1$
• C. $2K+5$
• D. $2K-5$

Hint:

Substituting $v = y - kx$ handles terms inside powers or functions in first-order differential equations smoothly.

Answer 1

The Correct Option is B

Step 1: Use a smart substitution.
The equation is $\dfrac{dy}{dx}=(y-kx)^2+K$. Let $v=y-kx$ to make it simpler.
Step 2: Differentiate the substitution.
$\dfrac{dv}{dx}=\dfrac{dy}{dx}-k$. So $\dfrac{dy}{dx}=\dfrac{dv}{dx}+k$.
Step 3: Plug in and simplify.
$\dfrac{dv}{dx}+k=v^2+k$, which gives the clean equation $\dfrac{dv}{dx}=v^2$.
Step 4: Separate and integrate.
$\dfrac{dv}{v^2}=dx$ gives $-\dfrac{1}{v}=x+c$, so $v=-\dfrac{1}{x+c}$.
Step 5: Use the starting values.
At $x=0,\ y=1$, we have $v=1-0=1$, so $1=-\dfrac{1}{c}$, giving $c=-1$. Hence $y-kx=\dfrac{1}{1-x}$, that is $y=kx+\dfrac{1}{1-x}$.
Step 6: Find $f(2)$.
\[ f(2)=k(2)+\frac{1}{1-2}=2k-1. \] \[ \boxed{2K-1} \]