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\(10y'' - $50$ y' + $65$ y = $0$ ext{.} ext{ , }y( $0$ )= $1$ . $5$ ext{. , }y'( $0$ )= $1$ . $5$ ext{.} ext{ }}

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Composition of linear functions: find a value

Full solution

Q. \(10y'' -

50

y' +

65

y =

0

ext{.} ext{ , }y(

0

)=

1

.

5

ext{. , }y'(

0

)=

1

.

5

ext{.} ext{ }}

Solve Characteristic Equation: First, we need to solve the characteristic equation for the differential equation $10r^2 - 50r + 65 = 0$ . Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , we get: $r = \frac{50 \pm \sqrt{50^2 - 4\cdot10\cdot65}}{2\cdot10}$ $r = \frac{50 \pm \sqrt{2500 - 2600}}{20}$ $r = \frac{50 \pm \sqrt{-100}}{20}$ $r = \frac{50 \pm 10i}{20}$ $r = 2.5 \pm 0.5i$
Write General Solution: Now we write the general solution for the differential equation using the roots found: $\newline$ $y(t) = e^{2.5t}(C_1\cos(0.5t) + C_2\sin(0.5t))$
Find $C_1$ and $C_2$ : Next, we need to find the values of $C_1$ and $C_2$ using the initial conditions. $\newline$ First, we'll use $y(0) = 1.5$ : $\newline$ $1.5 = e^{(2.5\cdot 0)}(C_1\cdot \cos(0\cdot 0.5) + C_2\cdot \sin(0\cdot 0.5))$ $\newline$ $1.5 = C_1$
Find Second Initial Condition: Now, we'll find $y'(t)$ to use the second initial condition $y'(0) = 1.5$ :
$y'(t) = 2.5e^{(2.5t)}(C1\cos(0.5t) + C2\sin(0.5t)) + e^{(2.5t)}(-0.5C1\sin(0.5t) + 0.5C2\cos(0.5t))$
$y'(0) = 2.5e^{0}(1.5\cos(0) + C2\sin(0)) + e^{0}(-0.5\cdot 1.5\sin(0) + 0.5C2\cos(0))$
$1.5 = 2.5\cdot 1.5 + 0.5C2$
$1.5 = 3.75 + 0.5C2$
$C2 = (1.5 - 3.75) / 0.5$
$C2 = -4.5 / 0.5$
$C2 = -9$

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