Question

Xdx+(y-5)dy=0 y(-4)=8​

Answer 1

Данное дифференциальное уравнение является уравнением с разделяющимися переменными

$(y-5)dy=-xdx\\ \\ \displaystyle \int (y-5)dy=-\int xdx~~~~\Leftrightarrow~~~~ \dfrac{y^2}{2}-5y=-\dfrac{x^2}{2}+C$

Получили общий интеграл. Найдем теперь частный интеграл, подставив начальные условия.

$\dfrac{8^2}{2}-5\cdot8=-\dfrac{(-4)^2}{2}+C\\ \\ C=0$

Частный интеграл: $\dfrac{y^2}{2}-5y=-\dfrac{x^2}{2}$

Answer 2

To solve the given first-order ordinary differential equation (ODE) $xdx + (y-5)dy = 0$ with the initial condition $y(-4) = 8$, we can follow the following steps:

Step 1: Separate the variables by moving all the terms involving $x$ to one side and the terms involving $y$ to the other side:

$xdx = -(y-5)dy$

Step 2: Integrate both sides of the equation with respect to their respective variables:

$\int xdx = -\int (y-5)dy$

Step 3: Evaluate the integrals:

$\frac{x^2}{2} = -\frac{y^2}{2} + 5y + C$

where $C$ is the constant of integration.

Step 4: Solve for $C$ using the initial condition $y(-4) = 8$:

$\frac{(-4)^2}{2} = -\frac{8^2}{2} + 5(8) + C$

$8 = -32 + 40 + C$

$8 = 8 + C$

$C = 0$

Step 5: Substitute the value of $C$ back into the equation:

$\frac{x^2}{2} = -\frac{y^2}{2} + 5y$

Step 6: Solve for $y$ in terms of $x$ by rearranging the equation:

$\frac{x^2}{2} + \frac{y^2}{2} - 5y = 0$

$\frac{x^2}{2} + \frac{y^2}{2} - 5y - \frac{y^2}{2} = 0$

$\frac{x^2}{2} - 5y = \frac{y^2}{2}$

$x^2 - 10y = y^2$

$x^2 = y^2 + 10y$

$x^2 = y(y + 10)$

Step 7: Substitute the initial condition $y(-4) = 8$ to find the particular solution:

$x^2 = 8(8 + 10)$

$x^2 = 8(18)$

$x^2 = 144$

$x = \pm 12$

Therefore, the particular solution to the given ODE with the initial condition $y(-4) = 8$ is $x = \pm 12$.