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azamat
2 years ago
11

In this problem,y = 1/(x2 + c)is a one-parameter family of solutions of the first-order DEy' + 2xy2 = 0.Find a solution of the f

irst-order IVP consisting of this differential equation and the given initial condition.
Mathematics
1 answer:
trasher [3.6K]2 years ago
6 0

Differentiate the given solution:

y=\dfrac1{x^2+C}\implies y'=-\dfrac{2x}{(x^2+C)^2}

Substitute y and y' into the ODE:

-\dfrac{2x}{(x^2+C)^2}+2x\left(\dfrac1{x^2+C}\right)^2=0

and it's easy to see the left side indeed reduces to 0.

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The correct interpretation of the probability P(X < 5) is; 0.8

<h3>How to Solve Probability?</h3>

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2 years ago
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Determine the value of variables a, b, and c that make each equation true.
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Corrected Question

Determine the values of a, b and c that make each equation true.

(x^a)^6=\dfrac{1}{x^{30}} \\\\(x^{-7})^{-4}=x^b\\\\(x^{-2})^c=x^{22}

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a=-5, b=28 and c=-11

Step-by-step explanation:

To solve for a,b and c, we apply the following laws of indices

\dfrac{1}{x^y}=x^{-y} \\\\(x^m)^n=x^{m X n}\\\\$If x^m=x^n,$ then m=n

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(x^a)^6=\dfrac{1}{x^{30}}\\x^{a*6}=x^{-30}\\6a=-30\\a=-5

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To solve for c

(x^{-2})^c=x^{22}\\x^{-2*c}=x^{22}\\-2c=22\\c=-11

4 0
3 years ago
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