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Lady bird [3.3K]
3 years ago
11

What’s the differential equation of

Mathematics
1 answer:
worty [1.4K]3 years ago
5 0

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 }

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