Solutions > Long Division with Remainder Calculator >

範囲 4ln(8x^2-7)-12sqrt(x+3)

Topic

Full pad

`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

lcm gcf factors long addition scientific

See All

area

asymptotes

critical points

derivative

domain

eigenvalues

eigenvectors

expand

extreme points

factor

implicit derivative

inflection points

intercepts

inverse

laplace

inverse laplace

partial fractions

range

slope

simplify

solve for

tangent

taylor

vertex

geometric test

alternating test

telescoping test

pseries test

root test

Steps Examples

Generated by AI

series solution d²xdt² −64x=0

Solution

x=c₁(1+32t²+512t⁴3 +… +4096t²ⁿ· 32(1)(2)(3)(4)… n(2n−1) +… )+c₂(t+32t³3 +512t⁵15 +… +4096t²ⁿ⁺¹· 32(2)(3)(4)(5)… n(2n+1) +… )

Show Steps

Hide Steps

Solution steps

One step at a time

d²xdt² −64x=0

Solve using series method: x=c₁(1+32t²+512t⁴3 +… +4096t²ⁿ· 32(1)(2)(3)(4)… n(2n−1) +… )+c₂(t+32t³3 +512t⁵15 +… +4096t²ⁿ⁺¹· 32(2)(3)(4)(5)… n(2n+1) +… )

x=c₁(1+32t²+512t⁴3 +… +4096t²ⁿ· 32(1)(2)(3)(4)… n(2n−1) +… )+c₂(t+32t³3 +512t⁵15 +… +4096t²ⁿ⁺¹· 32(2)(3)(4)(5)… n(2n+1) +… )