Verify by direct substitution that the given power series is a particular solution of the indicated differential equation. a. y = sigma^infinity_n = 0 (-1)^n/n! x^2n, y' + 2xy = 0
b. y = sigma^infinity_n = 0 (-1)^n x^2n, (1 + x^2)y' + 2xy = 0
c. y = sigma^infinity_n = 1 (-1)^n + 1/n x^n, (x + 1)y" + y' = 0
d. y = sigma^infinity_n = 0 (-1)^n/2^2n (n!)^2 x^2n, xy" + y' + xy = 0

1/2*qr*rs

step-by-step explanation:

a

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places