Respuesta :
Answer:
[tex]y=\frac{1}{x^{2}-6x+13 }[/tex]
Step-by-step explanation:
We have given,
[tex]\frac{dy}{dx}=y^{2}(6-2x)[/tex]
and initial condition [tex]x=3,\ y=\frac{1}{4}[/tex]
Now,
[tex]\frac{dy}{dx}=y^{2}(6-2x)[/tex]
Rearranging the variables, we get
[tex]\frac{dy}{y^{2} }=(6-2x).dx[/tex]
Applying integration both sides, we get
[tex]\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx[/tex]
⇒[tex]\frac{-1}{y} =6x-\frac{2x^{2} }{2}[/tex]
⇒ [tex]\frac{-1}{y}=6x-x^{2} +C[/tex]
Putting the initial condition (i.e., [tex]x=3,\ y=\frac{1}{4}[/tex]), we get
⇒ [tex]-4=6\times3-(3)^{2}+C[/tex]
⇒ [tex]-4=18-9+C[/tex]
∴ [tex]C=-13[/tex]
We have, [tex]\frac{-1}{y}=6x-x^{2} +C[/tex]
now putting the value of [tex]C[/tex] in above equation, we get
⇒ [tex]\frac{-1}{y}=6x-x^{2} -13[/tex]
⇒ [tex]\frac{1}{y}=-6x+x^{2} +13[/tex]
[tex]y=\frac{1}{x^{2}-6x+13 }[/tex]
