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stich3 [128]
2 years ago
10

Is it mathematically correct to say that:

Mathematics
1 answer:
Step2247 [10]2 years ago
8 0

Answer: Yes

Step-by-step explanation:

    Since the numerator is multiplying square root 13 by square root 2, we can break them apart.

\displaystyle \frac{\sqrt{13*2}}{\sqrt{13}} =\frac{\sqrt{13}*\sqrt{2}  }{\sqrt{13}}

   You are correct, \sqrt{13} over  \sqrt{13} is equal to one, so we can "cancel them out" of our expression.

\displaystyle \frac{\sqrt{13} }{\sqrt{13}} = 1

   This leaves us with: \sqrt{2}<em />

<em />

<em>We can also test this by using a calculator:</em>

<em />\frac{\sqrt{13*2}}{\sqrt{13}} \approx 1.41421356\\\sqrt{2} \approx 1.41421356\\1.41421356=1.41421356

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A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.
Reptile [31]

Answer:

a.P.I=\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)

b.G.S=C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}

Step-by-step explanation:

We are given that a linear differential equation

y''+2y=e^{3x}+x^3

We have to find the particular solution

P.I=\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}

P.I=\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}

P.I=\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3

P.I=\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0} (higher order terms can be neglected

P.I=\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)

b.Characteristics equation

D^2+2=0

D=\pm\sqrt2 i

C.F=C_1cos \sqrt2x+C_2 sin\sqrt2 x

G.S=C.F+P.I

G.S=C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)

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4 years ago
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