ksaimouli
  • ksaimouli
find particular solution y=f(x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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goformit100
  • goformit100
its easy
ksaimouli
  • ksaimouli
\[\frac{ dy }{ dx }=6x^2-x^2y\]
anonymous
  • anonymous
\[\Large \frac{dy}{dx}=(6-y)x^2 \] Or \[ \Large \frac{1}{6-y}dy=x^2dx \] For \(y(x) \neq 6\)

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More answers

ksaimouli
  • ksaimouli
\[\int\limits_{}^{}\frac{ 1 }{ 6-y }dy=x^2dx\]
anonymous
  • anonymous
Integrate both sides.
ksaimouli
  • ksaimouli
\[-lny=2x^3+c\]
ksaimouli
  • ksaimouli
@Spacelimbus
anonymous
  • anonymous
\[\Large - \ln |6-y(x)|=\frac{1}{3}x^3+C\prime  \]
ksaimouli
  • ksaimouli
they have given f(-1)=2
ksaimouli
  • ksaimouli
no calc
ksaimouli
  • ksaimouli
calculator
anonymous
  • anonymous
You know you have to solve it for y(x) if you can, of course - you could apply the initial conditions just now and solve for the constant. But I strongly recommend you to get the equation in explicit form if possible.
ksaimouli
  • ksaimouli
i got \[y(x)=ce ^{(x^3/x)}+6\]
anonymous
  • anonymous
Almost, try again. You should end up with: \[\Large y(x)=6-Ce^{-\frac{1}{3}x^3} \] As you can see, I said above that \(y(x) \neq 6\) which was important, otherwise I would have divided the differential equation through 0. This equation supports that statement, this equation only becomes 6 in the limit as x approaches infinity.
ksaimouli
  • ksaimouli
|dw:1360441531205:dw|
ksaimouli
  • ksaimouli
|dw:1360441583641:dw|
ksaimouli
  • ksaimouli
@Spacelimbus
ksaimouli
  • ksaimouli
i alway get these
anonymous
  • anonymous
Consider this equation: \[\Large e^{\ln2} = 2 \] but: \[\Large e^{-\ln2} = -2 ????? \] Rather view it like that: \[\Large e^{-\ln2}=\left(e^{\ln2}\right)^{-1}=\frac{1}{2} \]
ksaimouli
  • ksaimouli
okay
anonymous
  • anonymous
and then you should get the same expression as I do.

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