Question

Y"-3y'+2y=0 y=1 y'=1 x=0

Answer 1

Ответ: y = eˣ

Объяснение:

Answer 2

To find the solution to the differential equation $Y'' - 3Y' + 2Y = 0$ with the initial conditions $Y(0) = 1$ and $Y'(0) = 1$, we can solve it using the characteristic equation.

The characteristic equation is obtained by assuming a solution of the form $Y(x) = e^{rx}$ and plugging it into the differential equation. Substituting $Y(x) = e^{rx}$ into the differential equation gives:

$r^2 e^{rx} - 3r e^{rx} + 2e^{rx} = 0$

Factoring out the common term of $e^{rx}$ gives:

$e^{rx}(r^2 - 3r + 2) = 0$

For a non-trivial solution, we set the expression in parentheses equal to zero:

$r^2 - 3r + 2 = 0$

Now, we can solve this quadratic equation to find the values of $r$. Factoring the quadratic equation gives:

$(r - 2)(r - 1) = 0$

Setting each factor equal to zero, we have two solutions:

$r - 2 = 0$ or $r - 1 = 0$

$r_1 = 2$ or $r_2 = 1$

Since we have distinct real roots, the general solution of the differential equation is given by:

$Y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}$

where $C_1$ and $C_2$ are constants to be determined using the initial conditions.

To find the values of $C_1$ and $C_2$, we substitute the initial conditions $Y(0) = 1$ and $Y'(0) = 1$ into the general solution and solve the resulting equations.

Substituting $x = 0$ into the general solution gives:

$Y(0) = C_1 e^{r_1 \cdot 0} + C_2 e^{r_2 \cdot 0} = C_1 + C_2 = 1$

Substituting $x = 0$ into the derivative of the general solution gives:

$Y'(0) = r_1 C_1 e^{r_1 \cdot 0} + r_2 C_2 e^{r_2 \cdot 0} = r_1 C_1 + r_2 C_2 = 1$

Using $r_1 = 2$ and $r_2 = 1$, we can rewrite the above equations as a system of linear equations:

$C_1 + C_2 = 1$

$2C_1 + C_2 = 1$

Solving this system of equations, we find $C_1 = \frac{1}{3}$ and $C_2 = \frac{2}{3}$.

Therefore, the solution to the given differential equation with the initial conditions