Решение дифференциальных уравнений второго порядка
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.
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First integration:
\(y' = \int (\sin 4x + 2x - 3) dx = -\frac{1}{4}\cos 4x + x^2 - 3x + C_1\) -
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.
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First integration:
\(y' = \int (e^{5x} + \cos x - 2x^3) dx = \frac{1}{5}e^{5x} + \sin x - \frac{1}{2}x^4 + C_1\) -
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.
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First integration:
\(y' = \int (e^{2x+5} - 3x^2 + 5) dx = \frac{1}{2}e^{2x+5} - x^3 + 5x + C_1\) -
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.
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First integration:
\(y' = \int (\cos 5x + \sin 3x) dx = \frac{1}{5}\sin 5x - \frac{1}{3}\cos 3x + C_1\) -
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.
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First integration:
\(y' = \int (\cos(5x-3) + \sin 7x + \frac{1}{x^2}) dx = \frac{1}{5}\sin(5x-3) - \frac{1}{7}\cos 7x - \frac{1}{x} + C_1\) -
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!
