Explained: "Solve the initial value problem:16y″−40y′+25y=0Where the initial conditions..."

abreviatsjw

Answered question

2021-12-18

Solve the initial value problem:

16 y^{″} - 40 y^{'} + 25 y = 0

Where the initial conditions are:

y (0) = 3, y^{'} (0) = - \frac{9}{4}

Answer & Explanation

jean2098

Beginner2021-12-19Added 38 answers

Initial value problem
$16 y^{″} - 40 y^{'} + 25 y = 0$ We can find solution of the equation:
$16 D^{2} - 40 D + 25 = 0$ where $D = \frac{d}{d x}$
$16 D^{2} - 20 D - 20 D + 25 = 0$
$= 40 (40 - 5) - 5 (40 - 5) = 0$
$= (40 - 5) (40 - 5) = 0$
$= D = \frac{5}{4}; \frac{5}{4}$
root are real then solution of intial value problem
$y = (c_{1} x + c_{2}) e^{\frac{5}{4} x}$ ...(1)
Now differentiate then:
$y^{'} = \frac{5}{4} (c_{1} x + c_{2}) e^{\frac{5}{4} x} + c_{1} e^{\frac{5}{4}}$ ...(2)

Bubich13

Beginner2021-12-20Added 36 answers

Now we initial Conditions in equation 1 and 2

y (0) = (c_{1} x_{0} + c_{2}) e^{\frac{5}{4} x_{0}}

3 = c_{2} x_{1}

c_{2} = 3

y^{'} (0) = \frac{5}{4} (c_{1} x + c_{2}) e^{\frac{5}{4} x} + c_{1} e^{\frac{5}{4} x}

= - \frac{9}{4} = \frac{5}{4} (c_{1} x_{0} + c_{2}) e^{\frac{5}{4} x_{0}} + c_{1} e^{\frac{5}{4} x_{0}}

= - \frac{9}{4} = \frac{5}{4} \times 3 + 4

= - \frac{9}{4} = \frac{15}{4} + c_{1}

c_{1} = - \frac{9}{4} - \frac{15}{4}

c_{1} = \frac{- 9 - 15}{4}

c_{1} = - 6

Solution of the equation is:

y = (- 6 x + 3) e^{\frac{5}{4} x}

nick1337

Expert2021-12-28Added 777 answers

Good answers!

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