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aev [14]
2 years ago
13

8) (10) Solve the equation y''=x^2-y+3y' y(0)=1 and y'(0) =-1

Mathematics
1 answer:
defon2 years ago
5 0

y''-3y'+y=x^2

The corresponding homogeneous equation,

r^2-3r+1=0

has two roots at r=\dfrac{3\pm\sqrt5}2, so that the characteristic solution is

y_c=C_1e^{(3+\sqrt5)/2\,x}+C_2e^{(3-\sqrt5)/2\,x}

For the particular solution, assume one of the form

y_p=ax^2+bx+c

\implies{y_p}'=2ax+b

\implies{y_p}''=2a

Substituting y_p and its derivatives into the non-homogeneous ODE gives

2a-3(2ax+b)+(ax^2+bx+c)=x^2

ax^2+(-6a+b)x+(2a-3b+c)=x^2

\implies\begin{cases}a=1\\-6a+b=0\\2a-3b+c=0\end{cases}\implies a=1,b=6,c=16

Then the general solution to the ODE is

\boxed{y(x)=C_1e^{(3+\sqrt5)/2\,x}+C_2e^{(3-\sqrt5)/2\,x}+x^2+6x+16}

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1 + (72 ÷ (4 × 2)) =10

Step-by-step explanation:

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OverLord2011 [107]

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u>

\displaystyle \large{ {a}^{ - n}  =  \frac{1}{ {a}^{n} } }

Compare the terms.

\displaystyle \large{ {a}^{ - n}  =   {( - 2)}^{ - 3} }

Therefore, a = -2 and n = 3. From the law of exponent above, we receive:

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{ {( - 2)}^{ 3} } }

<u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>D</u><u>e</u><u>f</u><u>.</u> (For cubic)

\displaystyle \large{ {a}^{3}  = a \times a \times a }

Factor (-2)^3 out.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{( - 2) \times ( - 2) \times ( - 2)}}

(-2) • (-2) = 4 | Negative × Negative = Positive.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{4 \times ( - 2)}}

4 • (-2) = -8 | Negative Multiply Positive = Negative.

\displaystyle \large{ {( - 2)}^{ - 3}  =  \frac{1}{  - 8}}

If either denominator or numerator is in negative, it is the best to write in the middle or between numerator and denominators.

Hence,

\displaystyle \large \boxed{ {( - 2)}^{ - 3}  =  -  \frac{1}{  8}}

The answer is - 1 / 8

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Step-by-step explanation:

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