Question

Диф. уравнения 1го порядка

Answer 1

Ответ:

Решения дифференциальных уравнений:

5.

 $\boldsymbol{\boxed{C = \ln |x| + e^{\bigg(-\dfrac{y}{x} \bigg)}}}$

6.

$\boldsymbol{\boxed{\text{arctg} \dfrac{y}{x} - \dfrac{1}{2} \ln \bigg | \bigg(\dfrac{y}{x}\bigg)^{2} + 1 \bigg | - \ln|x| = C}}$

Пошаговое объяснение:

5.

$xy' = xe^{\bigg(\dfrac{y}{x} \bigg) } + y; x \neq 0$

$x\ \dfrac{dy}{dx} = xe^{\bigg(\dfrac{y}{x} \bigg) } + y$

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Замена: $y = tx \Longrightarrow t = \dfrac{y}{x} ; t = t(x)$

$\dfrac{dy}{dx} = \dfrac{d}{dx} \bigg(tx \bigg)= x \ \dfrac{dt}{dx} + t \dfrac{d}{dx} \bigg(x \bigg) = x \ \dfrac{dt}{dx} + t$

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$x\bigg(x \ \dfrac{dt}{dx} + t \bigg) = xe^{\bigg(\dfrac{tx}{x} \bigg) } + tx |:x$

$x \ \dfrac{dt}{dx} + t = e^{t} + t$

$x \ \dfrac{dt}{dx} = e^{t}$

$\dfrac{dt}{e^{t}} = \dfrac{dx}{x}$

$e^{-t}\, dt = \dfrac{dx}{x}$

$\displaystyle \int e^{-t}\, dt = \int \dfrac{dx}{x}$

$\displaystyle -\int e^{-t}\, d(-t) = \int \dfrac{dx}{x}$

$-e^{-t} + C = \ln |x|$

$C = \ln |x| + e^{-t}$

$\boxed{C = \ln |x| + e^{\bigg(-\dfrac{y}{x} \bigg)}}$ - общий интеграл дифференциального уравнения

6.

$y' = \dfrac{x+y}{x-y}$

$\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$

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Замена: $y = tx \Longrightarrow t = \dfrac{y}{x} ; t = t(x)$

$\dfrac{dy}{dx} = \dfrac{d}{dx} \bigg(tx \bigg)= x \ \dfrac{dt}{dx} + t \dfrac{d}{dx} \bigg(x \bigg) = x \ \dfrac{dt}{dx} + t$

$x = 0$ - не является решением дифференциального уравнения

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$x \ \dfrac{dt}{dx} + t = \dfrac{tx + x}{-tx+x}$

$x \ \dfrac{dt}{dx} + t = \dfrac{x(t +1)}{x(-t+1)}$

$x \ \dfrac{dt}{dx} = \dfrac{t +1}{1- t} -t$

$x \ \dfrac{dt}{dx} = \dfrac{t +1}{1- t} - \dfrac{t(1- t)}{1- t}$

$x \ \dfrac{dt}{dx} = \dfrac{t +1 - t(1-t)}{1- t}$

$x \ \dfrac{dt}{dx} = \dfrac{t +1 - t+t^{2}}{1- t}$

$x \ \dfrac{dt}{dx} = \dfrac{t^{2}+1}{1- t}$

$\dfrac{1-t}{t^{2}+1} \, dt = \dfrac{dx}{x}$

$\displaystyle \int \dfrac{1-t}{t^{2}+1} \, dt = \int \dfrac{dx}{x}$

а)

$\displaystyle \int \dfrac{1-t}{t^{2}+1} \, dt = \int \bigg( \dfrac{1}{t^{2}+1} - \dfrac{t}{t^{2}+1} \bigg ) \, dt = \int \dfrac{1}{t^{2}+1} \, dt - \int \dfrac{t}{t^{2}+1} \, dt =$

$\displaystyle = \int \dfrac{dt}{t^{2}+1} - \frac{1}{2} \int \dfrac{d(t^{2} + 1)}{(t^{2}+1)} = \text{arctg} \ t + C_{1} - \frac{1}{2} \ln |t^{2} + 1| + C_{2} =$

$= \text{arctg} \ t - \dfrac{1}{2} \ln |t^{2} + 1| + C_{3}$

б)

$\displaystyle \int \dfrac{dx}{x} = \ln|x| + C_{4}$

$\displaystyle \int \dfrac{1-t}{t^{2}+1} \, dt = \int \dfrac{dx}{x}$

$\text{arctg} \ t - \dfrac{1}{2} \ln |t^{2} + 1| + C_{3} = \ln|x| + C_{4}$

$\text{arctg} \ t - \dfrac{1}{2} \ln |t^{2} + 1| - \ln|x| = C_{4} - C_{3}$

$\boxed{\text{arctg} \dfrac{y}{x} - \dfrac{1}{2} \ln \bigg | \bigg(\dfrac{y}{x}\bigg)^{2} + 1 \bigg | - \ln|x| = C}$ - общий интеграл дифференциального уравнения