Question

Solve the autonomous differential equation. (independent variable is absent)
\[\frac{\mathrm d^2y}{\mathrm dx^2}+n^2y=0\]

Answer 1

\[y=A\cos nx + B\sin nx\]

Answer 2

im having trouble with the autonomus method

Answer 3

Is \(n\) just a constant?

Answer 4

yes

Answer 5

This is the way I was taught to do these. Let \(u=\frac{dy}{dx}\). Then
\[
\frac{d^2y}{dx^2} = \frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx} = y\frac{du}{dy}
\]and when you substitute than into the equation you get a first order ODE for \(u\) in terms of \(y\). Solve for \(u\).

Answer 6

Sorry, that should be \(u\frac{du}{dy}\) at the end there. I knew something didn't look right...

Answer 7

Possibly a better way to do it is to assume that \(e^{\lambda x}\) is a solution, which if you substitute into the equation, gives
\[\lambda^2 e^{\lambda x} + n^2e^{\lambda x} = 0
\]and of course you can divide out \(e^{\lambda x}\), so
\[\lambda^2 + n^2 = 0
\]so \(\lambda = \pm in\), and so your general solution is
\[
y=ae^{in} + be^{-in}
\]where \(a\) and \(b\) are arbitrary constants. You can put that in trig form using Euler's identity.

Answer 8

\[\newcommand \p \newcommand
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\p \den [3] {\frac{\mathrm d^#3 #1}{\mathrm d#2^#3}}
\begin{align} \den yx2+n^2y&=0 &\text{$x$-absent} \\
&&\text {let }\de yx&=u \\
&&\den yx2&=\de ux=\de uy\de yx=\de uyu \\
\de uyu+n^2y &=0 \\
u\de uy &=-n^2y \\
u\dd u &=-n^2y\dd y \\
\int u\dd uc &=-n^2\int y\dd y \\
\frac{u^2}2 &=-\frac{n^2y^2+c}2 \\
u^2 &=-n^2y^2+c \\
% \de yx &=\sqrt{}
% &=a\cos nx+b\sin nx
\end{align}\]

Answer 9

\[\frac{ d ^{2}y }{ dx ^{2} }+n ^{2}y=0\rightarrow \frac{ d ^{2}y }{ -n ^{2}y }=dx ^{2}\rightarrow \int\limits_{}^{}\int\limits_{}^{}\frac{ d ^{2y} }{ -n ^{2}y }=\int\limits_{}^{}\int\limits_{}^{}dx ^{2}\rightarrow
-\frac{ 1 }{ n ^{2} }(y \ln (y)-y)=\frac{ x ^{2} }{ 2 }+c _{1}x-c _{2}\]
figure out the y yourself

Answer 10

\[
\newcommand \p \newcommand
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\p \den [3] {\frac{\mathrm d^#3 #1}{\mathrm d#2^#3} }
\p \pa [2] {\frac{\partial #1}{\partial #2} }
\p \pan [3] {\frac{\partial^#3 #1}{\partial#2^#3} }
\p \pat [2] {\tfrac{\partial #1}{\partial #2} }
\begin{align} \den yx2+n^2y&=0 &\text{$x$-absent} \\
&&\text {let }\de yx &=p \\
&&\den yx2 &=\de px=\de py\de yx=p\de py \\
p\de py+n^2y &=0 \\
p\de py &=-n^2y \\
p\dd p &=-n^2y\dd y \\
\int p\dd p &=-n^2\int y\dd y \\
\frac{p^2}2&=\frac{-n^2y^2+c}2 \\
p^2 &=n^2(a^2-y^2) \\
p &=n\sqrt{a^2-y^2} \\
\de yx &=n\sqrt{a^2-y^2} \\
\frac{\dd y}{\sqrt{a^2-y^2}} &=n\dd x \\
\int\frac{\dd y}{\sqrt{a^2-y^2}} &=n\int\dd x \\
\arcsin \Big(\frac ya\Big) &=nx+b \\
y &=a\sin(nx+b) \\
y &=a\big(\sin(nx)\cos b
+\cos(nx)\sin b\big) \\
\\
y(x) &=A\sin (nx)+B\cos (nx)
\end{align}\]

Answer 11

That bit that goes off the edge
***\[\newcommand \p \newcommand
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\begin{align} && &\text{$x$-absent} \\
&&\text {let }\de yx &=p \\
&&\den yx2 &=\de px=\de py\de yx=p\de py \\
\end{align}\]

Answer 12

u could take it as a separable equation(the easiest way!)

Answer 13

seperable, but you have to take double integral ,

Answer 14

but still the easiest way

Answer 15

dx^2 is diffferent to d^2x and i get confused, i havent learnt that technique in class yet

Answer 16

it's just how u write it ! u must understand how they're different

Answer 17

it think the eaisiest way is to take the charestic equation

m^2+n^2=0

m=±in

y=Acos nx+Bsin nx

but that wasnt the technique i was trying

Answer 18

and also u could take it as a constant coefficient equation

Answer 19

this is a copy of your method (that dosent run off the page)\[\frac{ d ^{2}y }{ dx ^{2} }+n ^{2}y=0\\\rightarrow \frac{ d ^{2}y }{ -n ^{2}y }=dx ^{2}\\\rightarrow \int\limits_{}^{}\int\limits_{}^{}\frac{ d ^{2y} }{ -n ^{2}y }=\int\limits_{}^{}\int\limits_{}^{}dx ^{2}\\\rightarrow
-\frac{ 1 }{ n ^{2} }(y \ln (y)-y)\\=\frac{ x ^{2} }{ 2 }+c _{1}x-c _{2}\]

Answer 20

your technique is for complicated combination of y' and y"

Answer 21

x- absent and y- absent are thte first two types of DE's in the chapter on 2nd order DEs

this question was from the first problem set

Answer 22

lots of ways u can solve a differential equation but u must choose the best one.
u see I'm getting ready for my college entrance which is the hardest problems of differential equations will be presented in the exam that is why I was insisting on separable!

Answer 23

Yes, your right seperation of variables has got to be the most elementry technique,

Answer 24

i just haven't learnt it properlly in my class,

what is the difference between

\[\iint \mathrm d^2x\]
and
\[\iint \mathrm dz^2\]

Answer 25

what do u think of this ? what kind of equation is this?

Answer 26

\[\sin y\frac{\mathrm dy}{\mathrm dx}=\cos y (1-x\cos y)\]

Answer 27

in dz2 z must be the dependent variable otherwise u can't take double integral!
u know what I mean?

Answer 28

okay like i should of used y

Answer 29

go on

Answer 30

\[d(x ^{2}) is different from dx ^{2}\]

Answer 31

sorry

Answer 32

\[d(x ^{2})\] is different from \[dx ^{2}\]

Answer 33

u get it?

Answer 34

ok so i should have asked
the difference between \(\iint \mathrm d^2x\)
and \(\iint \mathrm d(y^2)\)

Answer 35

\[d ^{2}x =(dx)(dx)\]
but in \[∬d(y ^{2}) \] you're integrating with respect to \[y\]

Answer 36

sorry that is \[y ^{2}\]

Answer 37

you're a college student ? right?
what major?

Answer 38

\[\sin y\frac{\mathrm dy}{\mathrm dx}=\cos y (1-x\cos y)\\
\cos y (x\cos y-1)\mathrm dx+\sin y\mathrm dy=0\\
\\\,\\
M_x=\cos^2y\qquad N_y=\cos y\qquad \text{inexact}\\ \qquad \,\\
\frac{N_y-M_x}{M}=\frac{\cos y(1-\cos y)}{\cos^2y}=\sec y-1\]

Integrating factor \[R(y)=\int(\sec y-1)\mathrm d y\]

Answer 39

hmm its not so easy to that integral of sec y

Answer 40

if u take v=sec(y) it would become a simple first order equation

Answer 41

now just see how simple is that if u look closely to the equation?