Question

1)y'+4y=2 2)y"-4y+13=0 Пожалуйста, очень срочно!!!

Answer 1

1) $y'+4y-2=0;\ (y-\frac{1}{2})'=-4(y-\frac{1}{2});\ y-\frac{1}{2}=Ce^{-4x}.$

Ответ: $y=Ce^{-4x}+\frac{1}{2}$

2) $y''-4y+13=0;\ (y-\frac{13}{4})''=4(y-\frac{13}{4}); y-\frac{13}{4}=C_1e^{2x}+C_2e^{-2x}.$

Ответ: $y=C_1e^{2x}+C_2e^{-2x}+\frac{13}{4}$

Answer 2

Solution to the First Equation: y' + 4y = 2

To solve the first equation, y' + 4y = 2, we can use the method of integrating factors. The integrating factor is given by the exponential of the integral of the coefficient of y, which in this case is 4. Therefore, the integrating factor is e^(4x).

Multiplying both sides of the equation by the integrating factor, we get:

e^(4x) * (y' + 4y) = e^(4x) * 2

Simplifying the left side using the product rule of differentiation, we have:

(e^(4x) * y)' = 2e^(4x)

Integrating both sides with respect to x, we get:

∫(e^(4x) * y)' dx = ∫2e^(4x) dx

Using the fundamental theorem of calculus, the left side simplifies to:

e^(4x) * y = ∫2e^(4x) dx

Integrating the right side, we have:

e^(4x) * y = ∫2e^(4x) dx = (1/2) * e^(4x) + C1

Dividing both sides by e^(4x), we get the solution for y:

y = (1/2) + C1 * e^(-4x)

where C1 is the constant of integration.

Solution to the Second Equation: y" - 4y + 13 = 0

To solve the second equation, y" - 4y + 13 = 0, we can use the characteristic equation method. Assume a solution of the form y = e^(rx), where r is a constant.

Substituting this into the equation, we get:

(e^(rx))" - 4(e^(rx)) + 13 = 0

Differentiating twice, we have:

r^2 * e^(rx) - 4 * r * e^(rx) + 13 * e^(rx) = 0

Factoring out e^(rx), we get:

(e^(rx)) * (r^2 - 4r + 13) = 0

For a non-trivial solution, the quadratic equation r^2 - 4r + 13 = 0 must have complex roots. The discriminant of the quadratic equation is negative, indicating complex roots.

Using the quadratic formula, we find the roots:

r = (4 ± √(-48)) / 2

r = (4 ± 4√(-3)i) / 2

Simplifying, we have:

r = 2 ± 2√(-3)i

The general solution for y is then given by:

y = e^(2x) * (C2 * cos(2√3x) + C3 * sin(2√3x))

where C2 and C3 are constants determined by initial conditions.

Please note that the solutions provided are based on the given equations. If there are any additional constraints or initial conditions, they need to be considered to obtain specific solutions.