Math / Calculus

Question $\varepsilon x$ if $y_{1}(x)=e^{x}$ is a solk for $y^{\prime \prime}-2 y^{\prime}+y=0$ , find the general solu

Studdy Solution

STEP 1

1. The differential equation given is linear and homogeneous.
2. We are given one solution $y_1(x) = e^x$ to the differential equation.
3. The differential equation is of second order, so the general solution will involve two linearly independent solutions.

STEP 2

1. Verify that $y_1(x) = e^x$ is a solution.
2. Find a second linearly independent solution using the reduction of order method.
3. Construct the general solution using the two linearly independent solutions.

STEP 3

Verify that $y_1(x) = e^x$ is a solution to the differential equation $y'' - 2y' + y = 0$ .
Calculate the first derivative $y_1'(x)$ :
$y_1'(x) = \frac{d}{dx} e^x = e^x$
Calculate the second derivative $y_1''(x)$ :
$y_1''(x) = \frac{d^2}{dx^2} e^x = e^x$
Substitute $y_1(x)$ , $y_1'(x)$ , and $y_1''(x)$ into the differential equation:
$y_1'' - 2y_1' + y_1 = e^x - 2e^x + e^x = 0$
Since the left-hand side equals zero, $y_1(x) = e^x$ is indeed a solution.

STEP 4

Find a second linearly independent solution using the reduction of order method.
Assume a solution of the form $y_2(x) = v(x) e^x$ , where $v(x)$ is an unknown function.
Calculate the derivatives:
$y_2'(x) = v'(x) e^x + v(x) e^x$ $y_2''(x) = v''(x) e^x + 2v'(x) e^x + v(x) e^x$
Substitute into the differential equation:
$y_2'' - 2y_2' + y_2 = (v''(x) e^x + 2v'(x) e^x + v(x) e^x) - 2(v'(x) e^x + v(x) e^x) + v(x) e^x$
Simplify:
$= v''(x) e^x = 0$
Since $e^x \neq 0$ , we have $v''(x) = 0$ .
Integrate $v''(x) = 0$ to find $v'(x)$ :
$v'(x) = C_1$
Integrate again to find $v(x)$ :
$v(x) = C_1 x + C_2$
Thus, the second solution is:
$y_2(x) = (C_1 x + C_2) e^x$
Choose $C_1 = 1$ and $C_2 = 0$ for simplicity, so:
$y_2(x) = x e^x$

STEP 5

Construct the general solution using the two linearly independent solutions $y_1(x) = e^x$ and $y_2(x) = x e^x$ .
The general solution is:
$y(x) = C_1 e^x + C_2 x e^x$
where $C_1$ and $C_2$ are arbitrary constants.
The general solution to the differential equation is:
$\boxed{y(x) = C_1 e^x + C_2 x e^x}$

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