Question

4 d²y/dx²-12dy/dx+5y=0 please help me​

Answer 1

Ответ:

y=C₁e^(5x/2)+C₂e^(x/2)

Пошаговое объяснение:

4y''-12y'+5y=0;

однородное линейное дифуравнение второго порядка. Думать не надо. Совсем. Надо запомнить три варианта решения в зависимости от корней характеристического уравнения:

1. Характеристическое уравнение:

4λ²-12λ+5=0;  

2. Решаем:

D=144-80=64; √D=√64=8;

λ₁₂=(12±8)/8;  λ₁=5/2; λ₂=1/2;

3. Корни действительные, различные, значит:

y=C₁e^(5x/2)+C₂e^(x/2)

Answer 2

To solve the given differential equation, we can use the method of finding the characteristic equation and then finding the roots of the characteristic equation. Let's go through the steps:

Step 1: Write the differential equation

The given differential equation is: 4(d²y/dx²) - 12(dy/dx) + 5y = 0

Step 2: Find the characteristic equation

To find the characteristic equation, we assume a solution of the form y = e^(rx), where r is a constant. Substituting this into the differential equation, we get: 4(r²e^(rx)) - 12(re^(rx)) + 5e^(rx) = 0

Step 3: Simplify the equation

We can simplify the equation by factoring out e^(rx): e^(rx)(4r² - 12r + 5) = 0

Step 4: Solve the characteristic equation

For the equation to hold true, either e^(rx) = 0 or (4r² - 12r + 5) = 0.

Since e^(rx) is never equal to zero, we focus on solving the quadratic equation (4r² - 12r + 5) = 0.

Using the quadratic formula, we can find the roots of the quadratic equation: r = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 4, b = -12, and c = 5. Substituting these values into the quadratic formula, we get: r = (-(-12) ± √((-12)² - 4(4)(5))) / (2(4)) r = (12 ± √(144 - 80)) / 8 r = (12 ± √64) / 8 r = (12 ± 8) / 8

So the roots of the characteristic equation are: r₁ = (12 + 8) / 8 = 2 r₂ = (12 - 8) / 8 = 1/2

Step 5: Write the general solution

The general solution of the differential equation is given by: y = C₁e^(r₁x) + C₂e^(r₂x)

Substituting the values of r₁ and r₂, we get: y = C₁e^(2x) + C₂e^(1/2x)

where C₁ and C₂ are constants that can be determined from initial conditions or boundary conditions.

This is the general solution to the given differential equation.