Question

Solve the initial value problem for the differential equation given below: $y'(x) = 6x^2 - 4x^{-2}, y(1) = a$, where $a$ is a real number. a) $y(x) = 12x + 8x^{-3} - 20 + a$ b) $y(x) = 12x + \frac{4x^{-3}}{3} - \frac{40}{3} + a$ c) $y(x) = 2x^3 + 4x^{-1} - 6 + a$ d) $y(x) = 2x^3 - 4x^{-1} + 2 + a$ A function $f(x)$ has $2^{nd}$-order Taylor polynomial $p_2(x) = 2 + \frac{x-4}{4} - \frac{(x-4)^2}{64}$ centered at $a$.

          Solve the initial value problem for the differential equation given below:
$y'(x) = 6x^2 - 4x^{-2}, y(1) = a$,
where $a$ is a real number.
a) $y(x) = 12x + 8x^{-3} - 20 + a$
b) $y(x) = 12x + \frac{4x^{-3}}{3} - \frac{40}{3} + a$
c) $y(x) = 2x^3 + 4x^{-1} - 6 + a$
d) $y(x) = 2x^3 - 4x^{-1} + 2 + a$
A function $f(x)$ has $2^{nd}$-order Taylor polynomial $p_2(x) = 2 + \frac{x-4}{4} - \frac{(x-4)^2}{64}$ centered at $a$.

Solve the initial value problem for the differential equation given below:
y'(x) = 6x^2 - 4x^-2, y(1) = a,
where a is a real number.
a) y(x) = 12x + 8x^-3 - 20 + a
b) y(x) = 12x + (4x^-3)/(3) - (40)/(3) + a
c) y(x) = 2x^3 + 4x^-1 - 6 + a
d) y(x) = 2x^3 - 4x^-1 + 2 + a
A function f(x) has 2^nd-order Taylor polynomial p2(x) = 2 + (x-4)/(4) - ((x-4)^2)/(64) centered at a.

Added by Katherine B.

Differential Equations

Shepley L. Ross 3rd Edition

Chapter 2

Instant Answer

Solved by Expert Brent Burkett

Step 1

To do this, we can integrate both sides of the equation with respect to x: ∫y'(x) dx = ∫(62 - 4x^2) dx Integrating y'(x) gives us y(x) and integrating 62 - 4x^2 gives us 62x - (4/3)x^3 + C, where C is the constant of integration. So the general solution of the Show more…

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Solve the initial value problem for the differential equation given below: y'(x) = 62 - 4x^2, y(1) = a, where a is a real number. Oa) y(x) = 12x + 8x^3 - 20 + a Ob) 4x^3 - 40y = 12x + a^3 PO y(z) = 2x^3 + 4x - 1 - 6 + a Od) y(x) = 2x^3 - 4x - 1 + 2 + a A function f(x) has a 2nd-order Taylor polynomial p2(x) = 64 centered at a.