# DaniilGubajdullin
# Daniil Gubajdullin, 24.10.2001
## Task 1
$\begin{equation*}
\begin{cases}
y''y' = \frac{1}{2}\\
y(1) = (day)\\
y'(1) = (month)
\end{cases}
\end{equation*}$

$(day) = 24$
$(month) = 10$
$(year) = 2001$
### Solution:
$y''y' = \frac{1}{2}$ - second order nonlinear ordinary differential equation
1. Let $y' = p$, which gives $y'' = p'$, then integrate both sides with respect of x:
$\int p'p dx = \int\frac{1}{2} dx$
$~~~~~~~~~\Updownarrow$
$~~~\int p dp = \frac{1}{2}\int dx$
$~~~~~~~~~\Updownarrow$
$~~~~~~~~~\frac{p^2}{2} = \frac{x}{2} + c$, where $\textbf{c}$ - const and $\textbf{c}\in \mathbb{R}$
$~~~~~~~~~\Updownarrow$
$~~~~~~~~~~~p = \pm \sqrt{x+c}$

2. Substitute back for $y' = p$:
$y' = \pm \sqrt{x+c}$
$$a) Subsitute initial condition $y'(1) = (month) = 10$ into $y' = \sqrt{x+c}$:
$~~~~10 = \sqrt{c + 1}$
$~~~~~~~~~\Updownarrow$
$~~~~~~~c = 99$
b) Subsitute initial condition $y'(1) = (month) = 10$ into $y' = -\sqrt{x+c}$:
$~~~~10 = -\sqrt{1 + c}$
$~~~~$This equation has no solution.
3. Integrate both sides of $y' = \sqrt{x+99}$ with respect to x, $x \in [-99; +\infty)$:
$y = \int \sqrt{x+99} dx$
$~~~~~~~~~\Updownarrow$
$y = \frac{2}{3}(x+99)^{\frac{3}{2}} + c$, where $\textbf{c}$ - const and $\textbf{c}\in \mathbb{R}$
Subsitute initial condition $y(1) = (day) = 24$ into $y = \frac{2}{3}(x+99)^{\frac{3}{2}} + c$:
$~~~~24 = \frac{2}{3}(100)^{\frac{3}{2}} + c$
$~~~~~~~~~\Updownarrow$
$~~~~~~~c = 24 - \frac{2}{3}10^3$
$~~~~~~~~~\Updownarrow$
$~~~~~~~c = -\frac{1928}{3}$
4. Substitute $c = -\frac{1928}{3}$ into $y = \frac{2}{3}(x+99)^{\frac{3}{2}} + c$:
$y = \frac{2}{3}(x+99)^{\frac{3}{2}} - \frac{1928}{3}$
$~~~~~~~~~\Updownarrow$
$y = \frac{2}{3}((x+99)^{\frac{3}{2}} - 964)$

## Answer:
$$y = \frac{2}{3}((x+99)^{\frac{3}{2}} - 964)~~$$ on $$x \in [-99; +\infty)$$

## Task 2
$y'' + (month)y' -(day)y = 0$

$(day) = 24$
$(month) = 10$
$(year) = 2001$

### Solution:
$y'' + (month)y' -(day)y = 0$ - second order linear homogeneous ordinary differential equation with constant coefficients;
1. Let $y = e ^ {ax}$, where $a$ - const:
$(e ^ {ax})'' + 10 (e ^ {ax})' - 24 (e ^ {ax}) = 0$
$~~~~~~~~~\Updownarrow$
$a^2e ^ {ax} + 10 a e ^ {ax} - 24 e ^ {ax} = 0$
$~~~~~~~~~\Updownarrow$
$(a^2 + 10 a - 24) e ^ {ax} = 0$
$~~~~~~~~~\Updownarrow~~$ Since $e ^ {ax} \ne 0$ for any $a$
$a^2 + 10 a - 24 = 0$
$~~~~~~~~~\Updownarrow$
$(a - 2)(a + 12) = 0$
$~~~~~~~~~\Updownarrow$
$a = 2$ or $a = -12$
$~~~~~~~~~\Updownarrow$
$y_1 = c_1e^{2x};~~y_2 = c_2e^{-12x}$ $\Rightarrow$ $y = y_1 + y_2 = c_1e^{2x} + c_2e^{-12x}$, where $c_1$ and $c_2$ - const and $c_1, c_2 \in \mathbb{R}$
$y = c_1e^{2x} + c_2e^{-12x}$ - the most general solution on $x \in \mathbb{R}$, where $c_1, c_2$ - const and $c_1, c_2 \in \mathbb{R}$

## Answer:
$$y = c_1e^{2x} + c_2e^{-12x}~~$$