Question

bernoulli equation..it will take a min to type out everything...solve 2ty'-y=2ty^3cost

Answer 1

\[dy/dt-y/2t=y ^{3}cost\]
divide y^3 from both sides
\[y ^{-3}(dy/dt-y/(2t))=cost\]
n=3, P=1/2t for standard form dy/dt+P(t)y=Q(t)y^n
w=y^(1-n)=y^-2
dw/dt=-2y^-3dy/dt dy/dt=-y^3/2*dw/dt

substitue dy/dt
\[y ^{-3}(-y ^{3}dw/2dt-y/(2t))=cost\]
\[-dw/2dt-1/(y ^{2}2t)=cost\]

anything wrong so far? now integrating factor method...
\[\mu(t)=e ^{\int\limits_{}^{}1/(2t)dt}=t ^{1/2}\]
\[t ^{1/2}(-dw/2dt-w/(2t))=t ^{1/2} cost\] the w replaced 1/y^2
and now the part i'm stuck at...to find w, i have to integrate t^(1/2)cost...can't do it

Answer 2

multiply it thru by : y^-3

\[ 2t y^{-3} y' - y^{-2} = 2t cost\]

divide off the 2t

\[ y^{-3} y' - (2t)^{-1}y^{-2} = cost\]

substitute z = y^-2
z^-1/2 = y; \(\frac{-1}{2}z^{-3/2}z'=y'\)
z^3/2 = y^-3

\[ \frac{-1}{2}z^{3/2}*z^{-3/2} \ z' - (2t)^{-1}z = cost\]
\[ \frac{-1}{2}z' - (2t)^{-1}z = cost\]

multiply by -2

\[ z' + (t)^{-1}z = -2cost\]
and solve for z

Answer 3

ive always used a "z" to name my substitute with

Answer 4

when you multiply thru by that y^-n you automatically get you sub value of y^1-n

Answer 5

i see where i went wrong, getting rid of the 2 made the difference thanks!

Answer 6

youre welcome :)

Answer 7

i have a similar problem :(
y'-2y=y^(-1/2)cost
w=y^(3/2)
y'=2/3y^(-1/2)w'

w'-3w=3cost.

next is int factor method..
mu(t)=e^(3t)
w=(integral[e^(3t)3costdt]+C)/e^(3t)
integrate e^(3t)3cost
i must have done something wrong again

Answer 8

\[y^{1/2}(y'-2y = y^{-1/2}cost)\]
\[y^{1/2}y'-2y^{3/2} = cost\]
\begin{align}
z&=y^{3/2}\\
z^{2/3}&=y\implies &z^{1/3}=y^{1/2}\\
\frac{2}{3}z^{-1/3}z'&=y'
\end{align}
\[[z^{1/3}]\ [\frac{2}{3}z^{-1/3}z' ]-2[z] = cost\]
\[\frac{3}{2}(\frac{2}{3}z'-2z = cost)\]
\[z'-3z = \frac{3}{2}cost\]

Answer 9

\[e^{-3x}(z'-3z = \frac{3}{2}cost)\]
\[z'e^{-3x}-3e^{-3x}z = \frac{3}{2}e^{-3x}cost\]
\[-3e^{-3x}z = \frac{3}{2}\int e^{-3x}cost\ dt\]

Answer 10

stupid mistakes! thanks again, but i can't integrate that! well wolfram can but i have no idea what it did

Answer 11

its integratable by parts and it should come back to itself so that you can "undo" it