ksaimouli
find particular solution y=f(x
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goformit100
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its easy
ksaimouli
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\[\frac{ dy }{ dx }=6x^2-x^2y\]
Spacelimbus
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\[\Large \frac{dy}{dx}=(6-y)x^2 \]
Or
\[ \Large \frac{1}{6-y}dy=x^2dx \] For \(y(x) \neq 6\)
ksaimouli
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\[\int\limits_{}^{}\frac{ 1 }{ 6-y }dy=x^2dx\]
Spacelimbus
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Integrate both sides.
ksaimouli
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\[-lny=2x^3+c\]
ksaimouli
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@Spacelimbus
Spacelimbus
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\[\Large - \ln |6-y(x)|=\frac{1}{3}x^3+C\prime \]
ksaimouli
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they have given f(-1)=2
ksaimouli
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no calc
ksaimouli
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calculator
Spacelimbus
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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
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i got \[y(x)=ce ^{(x^3/x)}+6\]
Spacelimbus
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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
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|dw:1360441531205:dw|
ksaimouli
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|dw:1360441583641:dw|
ksaimouli
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@Spacelimbus
ksaimouli
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i alway get these
Spacelimbus
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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
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okay
Spacelimbus
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and then you should get the same expression as I do.