Common Sense = Equation

Calculus Level 2

There is a differential equation ${y}''+{b}{y}'+{c}{y}={0}$ and $\frac{{b}^2}{4}$ $\neq$ ${c}$ and the solution is $Ae^{{n}{t}}$ + $Be^{{m}{t}}$ . The equations ${r}+{q}=-{b}$ , ${r}{q}={c}$ , ${k_1}{r}={n}$ , and ${k_2}{q}={m}$ are true .Find $\frac{{k_1}-{2}{k_2}}{{8}}$

The answer is -0.125.

1 solution

Aaryan Vaishya
Sep 21, 2019

Before I start the solution let me give you an important lesson.11 year olds can and will fool you MUHUHAHUAHAHUAHUH. Anyways the equations tell us that the differential equation is easily factorable into (D-r)(D-q)y.We can set (D-q)y as z and get z'-rz = 0 so z = Ae^rt.So Ae^rt = y'-qy. Multiplying by e^-qt we have a product rule that we can integrate so Ae^(r-q)t+B = ye^-qt and finally multiplying back by e^qt we have y(t)=Ae^rt+Be^qt.Therefore r=n and q = m(or the other way around) so both k1 and k2 =1.(1-2)/8 = (-0.125).Note that the dividing during the integration just gets mixed in to the constant.

sry no time for latex

aaryan vaishya - 1 year, 8 months ago

Common Sense = Equation

The answer is -0.125.

1 solution

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