- #1
BOAS
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- 19
Hi,
I am struggling to find the solution to the following equation. I can't account for the exponential term, so clearly something is going wrong...
1. Homework Statement
Find the general solution to ##x' = tx + 6te^{-t^2}## where ##x(t)##.
[/B]
Consider ##x' - tx = 0##
##\frac{dx}{dt} = tx##
##\frac{dx}{t} = t dt##
Mod note: the above should be ##\frac{dx}{x} = t dt##
integrating and exponentiating gives ##x = Ae^{\int t dt}##. Let ##I = \int t dt##
##x = A e^{I}## and rearranging gives ##A = x e^{-I}##
Differentiating yields ##A' = x'e^{-I} - t x e^{-I}##
Factoring out the exponential term, and recognising that this is the same as the initial equation.
##A' = e^{-I}(x' - tx) = e^{-I}(6te^{-t^{2}})##
##A = \int e^{-I}(x' - tx) dt = \int e^{-I}(6te^{-t^{2}}) dt## which simplifies to
##A = \int e^{-\frac{3}{2} t^{2} + c} 6t dt##
Using the substitution ## u = - \frac{3}{2} t^{2} ## I find that
##A = - 2 \int e^{u} du = -2e^c e^{- \frac{3}{2} t^{2}} + c##
##x = e^I A = e^{- \frac{1}{2} t^2}(-2 e^{- \frac{3}{2} t^{2}} + c)##
This is not a solution to my differential equation. The ##6t e^{-t^2}## term is nowhere to be seen.
What am I doing wrong?
I am struggling to find the solution to the following equation. I can't account for the exponential term, so clearly something is going wrong...
1. Homework Statement
Find the general solution to ##x' = tx + 6te^{-t^2}## where ##x(t)##.
Homework Equations
The Attempt at a Solution
[/B]
Consider ##x' - tx = 0##
##\frac{dx}{dt} = tx##
##\frac{dx}{t} = t dt##
Mod note: the above should be ##\frac{dx}{x} = t dt##
integrating and exponentiating gives ##x = Ae^{\int t dt}##. Let ##I = \int t dt##
##x = A e^{I}## and rearranging gives ##A = x e^{-I}##
Differentiating yields ##A' = x'e^{-I} - t x e^{-I}##
Factoring out the exponential term, and recognising that this is the same as the initial equation.
##A' = e^{-I}(x' - tx) = e^{-I}(6te^{-t^{2}})##
##A = \int e^{-I}(x' - tx) dt = \int e^{-I}(6te^{-t^{2}}) dt## which simplifies to
##A = \int e^{-\frac{3}{2} t^{2} + c} 6t dt##
Using the substitution ## u = - \frac{3}{2} t^{2} ## I find that
##A = - 2 \int e^{u} du = -2e^c e^{- \frac{3}{2} t^{2}} + c##
##x = e^I A = e^{- \frac{1}{2} t^2}(-2 e^{- \frac{3}{2} t^{2}} + c)##
This is not a solution to my differential equation. The ##6t e^{-t^2}## term is nowhere to be seen.
What am I doing wrong?
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