Question

What’s the differential equation of

Answer 1

Answer:

$y=\frac{1}{x^{2}-6x+13 }$

Step-by-step explanation:

We have given,

                        $\frac{dy}{dx}=y^{2}(6-2x)$

and initial condition $x=3,\ y=\frac{1}{4}$

Now,

$\frac{dy}{dx}=y^{2}(6-2x)$

Rearranging the variables, we get

$\frac{dy}{y^{2} }=(6-2x).dx$

Applying integration both sides, we get

$\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx$

⇒$\frac{-1}{y} =6x-\frac{2x^{2} }{2}$

⇒ $\frac{-1}{y}=6x-x^{2} +C$                  

Putting the initial condition (i.e., $x=3,\ y=\frac{1}{4}$), we get

⇒ $-4=6\times3-(3)^{2}+C$

⇒ $-4=18-9+C$

∴ $C=-13$

We have,  $\frac{-1}{y}=6x-x^{2} +C$    

now putting the value of $C$ in above equation, we get

⇒ $\frac{-1}{y}=6x-x^{2} -13$

⇒  $\frac{1}{y}=-6x+x^{2} +13$

$y=\frac{1}{x^{2}-6x+13 }$