Question

Y''+y'-6y=0 если y=3 y'=1 при x=o

Answer 1

$y''+y-6y=0\\k^2+k-6=0\\k=2;k=-3\\y=C_1e^{-3x}+C_2e^{2x}$

Answer 2

To solve the given second-order ordinary differential equation (ODE) $Y'' + Y' - 6Y = 0$ with the initial conditions $Y(0) = 3$ and $Y'(0) = 1$, we can follow these steps:

Step 1: Find the general solution of the ODE. Step 2: Use the initial conditions to determine the specific solution.

Let's proceed with the calculations:

Step 1: Finding the general solution of the ODE. The characteristic equation for the ODE $Y'' + Y' - 6Y = 0$ is obtained by assuming $Y = e^{rx}$ and solving for $r$:

$r^2 + r - 6 = 0$

This quadratic equation can be factored as:

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

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

The general solution of the ODE is a linear combination of these roots:

$Y(x) = C_1e^{-3x} + C_2e^{2x}$

where $C_1$ and $C_2$ are arbitrary constants.

Step 2: Using initial conditions to find the specific solution.

Given $Y(0) = 3$ and $Y'(0) = 1$, we can use these initial conditions to find the values of $C_1$ and $C_2$:

$Y(0) = C_1e^{0} + C_2e^{0} = C_1 + C_2 = 3$

$Y'(x) = -3C_1e^{-3x} + 2C_2e^{2x}$

$Y'(0) = -3C_1e^{0} + 2C_2e^{0} = -3C_1 + 2C_2 = 1$

Now we have a system of equations:

$C_1 + C_2 = 3$

$-3C_1 + 2C_2 = 1$

Solving this system, we get $C_1 = 1$ and $C_2 = 2$.

Thus, the specific solution to the initial value problem is:

$Y(x) = 1e^{-3x} + 2e^{2x} = e^{-3x} + 2e^{2x}$

That's the final solution for the given initial value problem.