Solve the initial value problem dydx=(x−2)(y−10),y(0)=5dydx=(x−2)(y−10),y(0)=5.

Ответ:

JeonTaekook

03.01.2022

The ODE is separable:

$\dfrac{\mathrm dy}{\mathrm dx}=(x-2)(y-10)\implies\dfrac{\mathrm dy}{y-10}=(x-2)\,\mathrm dx$

Integrating both sides gives

$\ln|y-10|=\dfrac{(x-2)^2}2+C$

$\implies y-10=e^{(x-2)^2/2+C}=Ce^{(x-2)^2/2}$

$\implies y=10+Ce^{(x-2)^2/2}$

Given that y(0)=5 , we find

$5=10+Ce^{(-2)^2/2}\implies-5=Ce^2\implies C=-5e^{-2}$

so that

$\boxed{y(x)=10-5e^{(x-2)^2/2-2}}$

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14.12.2022

(2√5 + 3(√7))^2
(2√5 + 3(√7))(2√5 + 3(√7))
4*5 + 6√35 + 6√35 + 9*7
20 + 12√35 + 63
20 + 63 + 12√35
83 + 12√35

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