Question

1. xy' + y + xe^(-x^(2)) = 0 2. (x + 2y)dx + 2xdy = 03. y = y' ln y4. y" + 4y' + 4y = 05. y" + 10y' + 34y = -9e^(-5x)6. y" + 4y = 3cosx

Answer 1

$1) xy'+y+xe^{-x^2}=0$
Вычтем  $e^{-x^2}x$ с обеих сторон и разделим на  $x$:
$y'+ \frac{y}{x}= -e^{-x^2}$
Допустим, μ= $e^{ \int\limits { \frac{1}{x} } \, dx }=x$
Умножим обе части уравнения на μ:
$xy'+y=-e^{-x^2}x$
Замена:  $1=x'$:
$xy'+x'y=-e^{-x^2}x$
$(xy)'=-e^{-x^2}x$
$\int\limits {(xy)'} \, dx = \int\limits {-e^{-x^2}x} \, dx$
$xy= \frac{e^{-x^2}}{2}+C$
$y= \frac{ \frac{e^{-x^2}}{2}+C }{x}$

$2) (x+2y)dx+2xdy=0$
$2y+2xy'+x=0$
$y'+ \frac{y}{x} =- \frac{1}{2}$
Допустим, μ= $e^{ \int\limits { \frac{1}{x} } \, dx }=x$
$xy'+y=- \frac{x}{2}$
Замена:  $1=x'$
$xy'+x'y=- \frac{x}{2}$
$(xy)'=- \frac{x}{2}$
$\int\limits {(xy)'} \, dx = \int\limits {- \frac{x}{2} } \, dx$
$xy=- \frac{x^2}{4} +C$
$y=- \frac{x}{4} + \frac{C}{x}$

$3) y=y'ln y$
$y'= \frac{y}{lny}$
$\frac{ln(y)y'}{y} =1$
$\int\limits { \frac{ln(y)y'}{y} } \, dx = \int\limits {} \, dx$
$\frac{1}{2} ln^2(y)=x+C$
$y_1=e^{- \sqrt{2} \sqrt{x+C} }$
$y_2=e^{ \sqrt{2} \sqrt{x+C} }$

$4) y''+4y'+4y=0$;
Решим, как однородное линейное уравнение:
Допустим, решение будет решение будет пропорционально e^(λx) для некоторой константы λ.
Заменим y=e^(λx) в дифференциальное уравнение:
(e^(λx))''+4(e^(λx))'+4e^(λx)=0;
Заменим (e^(λx))''=λ²e^(λx)   и (e^(λx))'=λe^(λx):
λ²e^(λx)+4λe^(λx)+4e^(λx)=0;
(λ²+4λ+4)e^(λx)=0;
Т.к. e^(λx)≠0 для любого конечного λ, нули должны исходить от многочлена:
λ²+4λ+4=0;
(λ+2)²=0;
λ=-2   λ=-2;
Кратность корня  λ=-2 - это 2, делающее $y_1=Ce^{-2x}$,  $y_2=Ce^{-2x}x$ в качестве решения, где C -произвольная константа.
Ответ:  $y=y_1+y_2=C_1e^{-2x}+C_2e^{-2x}x$

Answer 2

1. To solve the differential equation xy' + y + xe^(-x^2) = 0:

Let's first rearrange the equation to isolate the derivative term:

xy' + y = -xe^(-x^2)

Now, we can rewrite the equation in a form suitable for the integrating factor method. The integrating factor is given by:

IF = e^(∫(1/x) dx) = e^(ln|x|) = |x|

Multiplying both sides of the equation by the integrating factor:

|x|xy' + |x|y = -|x|xe^(-x^2)

The left side can be rewritten using the product rule:

(d/dx)(|x|y) = -|x|xe^(-x^2)

Integrating both sides with respect to x:

∫(d/dx)(|x|y) dx = ∫(-|x|xe^(-x^2)) dx

Integrating the right side:

|x|y = ∫(-xe^(-x^2)) dx = (-1/2)e^(-x^2) + C

Now, we can solve for y:

y = (-1/2)(e^(-x^2) + C)/|x|

Therefore, the solution to the differential equation is:

y = (-1/2)(e^(-x^2) + C)/|x|, where C is a constant.

2. To solve the differential equation (x + 2y)dx + 2xdy = 0:

Let's rearrange the equation:

(x + 2y)dx = -2xdy

Now, we can rewrite it in a form suitable for separation of variables:

(x + 2y)dx + 2xdy = 0

Dividing throughout by x:

(1 + 2y/x)dx + 2dy = 0

Now, we can separate the variables and integrate:

∫(1 + 2y/x) dx + ∫2dy = 0

Integrating:

x + 2y ln|x| + 2y = C

Rearranging the equation:

x + 2y(1 + ln|x|) = C

Therefore, the solution to the differential equation is:

x + 2y(1 + ln|x|) = C, where C is a constant.

3. To solve the differential equation y = y' ln y:

Separate the variables:

(1/ln y) dy = dx

Integrating both sides:

∫(1/ln y) dy = ∫dx

This integral can be solved using a substitution. Let u = ln y, then du = (1/y)dy:

∫(1/u) du = ∫dx

ln|u| = x + C

Substituting back u = ln y:

ln|ln y| = x + C

Therefore, the solution to the differential equation is:

ln|ln y| = x + C, where C is a constant.

4. To solve the differential equation y" + 4y' + 4y = 0:

The characteristic equation for this second-order linear homogeneous differential equation is:

r^2 + 4r + 4 = 0

Solving the quadratic equation, we find that it has a repeated root at -2:

(r + 2)^2 = 0

Therefore, the general solution to the differential equation is:

y(x) = (c1 + c2x)e^(-2x), where c1 and c2 are constants.

5. To solve the differential equation y" + 10