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Orlov [11]
3 years ago
7

The method of Problem 20 can be extended to second order equations with variable coefficients. If y1 is a known nonvanishing sol

ution of y′′ + p(t)y′ + q(t)y = 0, show that a second solution y2 satisfies (y2 /y1 )′ = W (y1 , y2 )/y21 , where W (y1 , y2 ) is the Wronskian of y1 and y2 . Then use Abel’s formula [Eq. (23) of Section 3.2] to determine y2 .
Mathematics
1 answer:
lions [1.4K]3 years ago
7 0

Answer:24y

Step-by-step explanation:

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Five and three sixths minus two and one third
Daniel [21]

Answer:

19/6

Step-by-step explanation:

5 3/6 - 2 1/3 = 19/6

8 0
3 years ago
When a number is added to six times its square, the result is twelve. Find the number.
7nadin3 [17]
Let's represent our number as x.
x is being added to 6 * x², and it equals 12
x + 6x^{2}=12\\6x^{2}+x-12=0
Factor:
6x^{2}+x-12=0\\(2x+3)(3x-4)=0
Solve for x:
(2x+3)=0\\2x=-3\\x=\frac{-3}{2}\\\\(3x-4)=0\\3x=4\\x=\frac{4}{3}

Therefore, x = -3/2 and 4/3.
6 0
3 years ago
Read 2 more answers
Help please! I’m confused
Paul [167]

Answer:

28 degrees

Step-by-step explanation:

hi! since m<ABC is 175 degrees and m<YBC is 147 degrees, we can subtract these to find m<ABY.

175-147= 28 degrees

3 0
3 years ago
Ericka decided to compare her observation to the average annual trend, which shows the water rising 1.8 mm/year. Remember, she u
Alex787 [66]

Answer with explanation:

Presume that, initial water level to be, 0 mm, at a time t=0.

And water level is rising at the rate of 1.8 \frac{\text{mm}}{\text{year}}.

 After 6.2 years level of water will be =1.8 × 6.2

                                                           =11.16 mm

→Averge, that is increase in water level after 6.2 years is given by

                     =\frac{W_{2}-W_{1}}{t_{2}-t_{1}}\\\\=\frac{11.16-0}{6.2-0}\\\\=\frac{11.16}{6.2}\\\\=1.8 \frac{\text{mm}}{\text{year}}

→→Rate of Depreciating of water level is given by = Negative of Increase of water level=-1.8 \frac{\text{mm}}{\text{year}}

5 0
3 years ago
Read 2 more answers
Express p in terms of q if
Irina-Kira [14]

p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)

q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}

8 0
3 years ago
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