Question

Найти общее решение линейного дифференциального уравнения. Сделать проверку. y"-10y' + 34y = 0

Answer 1

Я решил на листочке, проверку сделай сам ===>>

Answer 2

Finding the General Solution of the Linear Differential Equation

To find the general solution of the given linear differential equation y"-10y'+34y=0, we can use the characteristic equation method. The characteristic equation for this equation is obtained by assuming a solution of the form y=e^(rx), where r is a constant. Substituting this into the differential equation, we get:

r^2e^(rx) - 10re^(rx) + 34e^(rx) = 0

Factoring out e^(rx), we have:

e^(rx)(r^2 - 10r + 34) = 0

For this equation to hold true, either e^(rx) = 0 or r^2 - 10r + 34 = 0.

Since e^(rx) is never equal to zero, we focus on solving the quadratic equation:

r^2 - 10r + 34 = 0

Using the quadratic formula, we find the roots of this equation:

r = (10 ± √(10^2 - 4*1*34)) / 2

Simplifying further, we have:

r = (10 ± √(-24)) / 2

Since the discriminant is negative, the roots are complex conjugates. Let's denote them as r1 = a + bi and r2 = a - bi, where a and b are real numbers.

The general solution of the differential equation is then given by:

y = C1e^(r1x) + C2e^(r2x)

where C1 and C2 are constants to be determined.

Verification of the Solution

To verify the solution, we can substitute it back into the original differential equation and check if it satisfies the equation.

Substituting y = C1e^(r1x) + C2e^(r2x) into the differential equation y"-10y'+34y=0, we get:

[(C1r1^2e^(r1x) + C2r2^2e^(r2x))] - 10[(C1r1e^(r1x) + C2r2e^(r2x))] + 34(C1e^(r1x) + C2e^(r2x)) = 0

Simplifying further, we have:

C1(r1^2 - 10r1 + 34)e^(r1x) + C2(r2^2 - 10r2 + 34)e^(r2x) = 0

Since r1 and r2 are the roots of the characteristic equation, we know that r1^2 - 10r1 + 34 = 0 and r2^2 - 10r2 + 34 = 0. Therefore, the equation simplifies to:

**C1(0)e^(r1x) + C2(0)e^(r