Okay, let's solve these differential equations one by one. I will provide step-by-step solutions with explanations in Markdown format.

Задание 1

$y'' = \sin 4x + 2x - 3$

To find $y$, we need to integrate twice.

1. First integration:
   $y' = \int (\sin 4x + 2x - 3) dx = -\frac{1}{4}\cos 4x + x^2 - 3x + C_1$

2. Second integration:
   $y = \int (-\frac{1}{4}\cos 4x + x^2 - 3x + C_1) dx = -\frac{1}{16}\sin 4x + \frac{1}{3}x^3 - \frac{3}{2}x^2 + C_1x + C_2$

Answer: $y = -\frac{1}{16}\sin 4x + \frac{1}{3}x^3 - \frac{3}{2}x^2 + C_1x + C_2$

Задание 2

$y'' = e^{5x} + \cos x - 2x^3$

To find $y$, we need to integrate twice.

1. First integration:
   $y' = \int (e^{5x} + \cos x - 2x^3) dx = \frac{1}{5}e^{5x} + \sin x - \frac{1}{2}x^4 + C_1$

2. Second integration:
   $y = \int (\frac{1}{5}e^{5x} + \sin x - \frac{1}{2}x^4 + C_1) dx = \frac{1}{25}e^{5x} - \cos x - \frac{1}{10}x^5 + C_1x + C_2$

Answer: $y = \frac{1}{25}e^{5x} - \cos x - \frac{1}{10}x^5 + C_1x + C_2$

Задание 3

$y'' = e^{2x+5} - 3x^2 + 5$

To find $y$, we need to integrate twice.

1. First integration:
   $y' = \int (e^{2x+5} - 3x^2 + 5) dx = \frac{1}{2}e^{2x+5} - x^3 + 5x + C_1$

2. Second integration:
   $y = \int (\frac{1}{2}e^{2x+5} - x^3 + 5x + C_1) dx = \frac{1}{4}e^{2x+5} - \frac{1}{4}x^4 + \frac{5}{2}x^2 + C_1x + C_2$

Answer: $y = \frac{1}{4}e^{2x+5} - \frac{1}{4}x^4 + \frac{5}{2}x^2 + C_1x + C_2$

Ассистент: Okay, let's continue solving the differential equations.

Задание 4

$y'' = \cos 5x + \sin 3x$

To find $y$, we need to integrate twice.

1. First integration:
   $y' = \int (\cos 5x + \sin 3x) dx = \frac{1}{5}\sin 5x - \frac{1}{3}\cos 3x + C_1$

2. Second integration:
   $y = \int (\frac{1}{5}\sin 5x - \frac{1}{3}\cos 3x + C_1) dx = -\frac{1}{25}\cos 5x - \frac{1}{9}\sin 3x + C_1x + C_2$

Answer: $y = -\frac{1}{25}\cos 5x - \frac{1}{9}\sin 3x + C_1x + C_2$

Задание 5

$y'' = \cos(5x-3) + \sin 7x + \frac{1}{x^2}$

To find $y$, we need to integrate twice.

1. First integration:
   $y' = \int (\cos(5x-3) + \sin 7x + \frac{1}{x^2}) dx = \frac{1}{5}\sin(5x-3) - \frac{1}{7}\cos 7x - \frac{1}{x} + C_1$

2. Second integration:
   $y = \int (\frac{1}{5}\sin(5x-3) - \frac{1}{7}\cos 7x - \frac{1}{x} + C_1) dx = -\frac{1}{25}\cos(5x-3) - \frac{1}{49}\sin 7x - \ln|x| + C_1x + C_2$

Answer: $y = -\frac{1}{25}\cos(5x-3) - \frac{1}{49}\sin 7x - \ln|x| + C_1x + C_2$

Задание 6

$xy'' = 1 + y'$

Let $v = y'$, then $v' = y''$. The equation becomes:

$xv' = 1 + v$

$\frac{dv}{dx} = \frac{1+v}{x}$

Separating variables:

$\frac{dv}{1+v} = \frac{dx}{x}$

Integrating both sides:

$\int \frac{dv}{1+v} = \int \frac{dx}{x}$

$\ln|1+v| = \ln|x| + C_1$

$1+v = e^{\ln|x| + C_1} = e^{C_1}x = Ax$ (where $A = e^{C_1}$)

$v = Ax - 1$

Since $v = y'$:

$y' = Ax - 1$

Integrating again:

$y = \int (Ax - 1) dx = \frac{1}{2}Ax^2 - x + C_2$

Answer: $y = \frac{1}{2}Ax^2 - x + C_2$

Ассистент: Okay, let's continue solving the differential equations.

Задание 7

$y''x\ln x = y'$

Let $v = y'$, then $v' = y''$. The equation becomes:

$v'x\ln x = v$

$\frac{dv}{dx} = \frac{v}{x\ln x}$

Separating variables:

$\frac{dv}{v} = \frac{dx}{x\ln x}$

Integrating both sides:

$\int \frac{dv}{v} = \int \frac{dx}{x\ln x}$

$\ln|v| = \int \frac{d(\ln x)}{\ln x} = \ln|\ln x| + C_1$

$v = e^{\ln|\ln x| + C_1} = e^{C_1}\ln x = A\ln x$ (where $A = e^{C_1}$)

Since $v = y'$:

$y' = A\ln x$

Integrating again:

$y = \int A\ln x dx = A\int \ln x dx$

Using integration by parts: $\int u dv = uv - \int v du$

Let $u = \ln x$, $dv = dx$, then $du = \frac{1}{x}dx$, $v = x$

$\int \ln x dx = x\ln x - \int x \cdot \frac{1}{x} dx = x\ln x - \int dx = x\ln x - x + C_2$

$y = A(x\ln x - x) + C_2$

Answer: $y = A(x\ln x - x) + C_2$

All tasks are completed!