Question

Y'' + y' – 6y=0; y0 = 3, x0 =0, y'0 =1

Answer 1

Пошаговое объяснение:

Получилось так.

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Answer 2

To solve the second-order linear homogeneous differential equation $y'' + y' - 6y = 0$ with the given initial conditions $y(0) = 3$ and $y'(0) = 1$, you can follow these steps:

Step 1: Find the characteristic equation The characteristic equation for this differential equation is obtained by replacing $y''$ with $r^2$, $y'$ with $r$, and $y$ with $1$:

$r^2 + r - 6 = 0$

Now, you need to solve this quadratic equation for $r$:

$r^2 + r - 6 = 0$

This equation can be factored as:

$(r + 3)(r - 2) = 0$

Setting each factor equal to zero gives two possible values for $r$:

1. $r + 3 = 0$ => $r = -3$
2. $r - 2 = 0$ => $r = 2$

So, the characteristic roots are $r_1 = -3$ and $r_2 = 2$.

Step 2: Write down the general solution The general solution of the differential equation is given by:

$y(x) = c1 * e^(r1 * x) + c2 * e^(r2 * x)$

Where $c1$ and $c2$ are constants to be determined and $r1$ and $r2$ are the characteristic roots we found earlier.

So, the general solution is:

$y(x) = c1 * e^(-3 * x) + c2 * e^(2 * x)$

Step 3: Use initial conditions to find the constants Now, you can use the initial conditions $y(0) = 3$ and $y'(0) = 1$ to find the values of $c1$ and $c2$.

1. $y(0) = 3$:

3 &= c1 * e^(-3 * 0) + c2 * e^(2 * 0) \\ 3 &= c1 + c2 \end{align*}\] 2. \(y'(0) = 1\): To find \(y'(x)\), differentiate the general solution: \[y'(x) = -3c1 * e^(-3 * x) + 2c2 * e^(2 * x)\] Now, evaluate \(y'(x)\) at \(x = 0\): \[1 = -3c1 * e^(-3 * 0) + 2c2 * e^(2 * 0) = -3c1 + 2c2\] Now, you have a system of two equations with two unknowns: \[\begin{align*} 3 &= c1 + c2 \\ 1 &= -3c1 + 2c2 \end{align*}\] You can solve this system of equations to find the values of \(c1\) and \(c2\). Solving the system, you'll find: \[c1 = 1\] \[c2 = 2\] So, the particular solution to the initial value problem is: \[y(x) = e^(-3 * x) + 2 * e^(2 * x)\]