Differential equation with initial conditions

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Apr 2010
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Find the unique function \(\displaystyle y(x)\) satisfying the differential equation with initial condition,

\(\displaystyle \frac{dy}{dx}=x^2 y\), \(\displaystyle y(1)=1\)
 
Jul 2007
894
298
New Orleans
Find the unique function \(\displaystyle y(x)\) satisfying the differential equation with initial condition,

\(\displaystyle \frac{dy}{dx}=x^2 y\), \(\displaystyle y(1)=1\)
\(\displaystyle \int \frac{dy}{y} = \int x^2 dx\)

\(\displaystyle \ln{|y|} = \frac{x^3}{3} + C\)

Can you take it fron here?
 
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Apr 2010
18
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I'm guessing I have to solve for y in terms of x, let y = 1 and x = 1, then solve for C?

This is what I did:

\(\displaystyle y = e^\frac{x^3}{3} + C\)

Since \(\displaystyle y(1)=1\),

\(\displaystyle 1=e^\frac{1}{3} + C\)

\(\displaystyle C=1 - e^\frac{1}{3}\)

For some reason this doesn't feel right.

Then the original equation should be:

\(\displaystyle y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}\)
 
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Jul 2007
894
298
New Orleans
I'm guessing I have to solve for y in terms of x, let y = 1 and x = 1, then solve for C?

This is what I did:

\(\displaystyle y = e^\frac{x^3}{3} + C\)

Since \(\displaystyle y(1)=1\),

\(\displaystyle 1=e^\frac{1}{3} + C\)

\(\displaystyle C=1 - e^\frac{1}{3}\)

For some reason this doesn't feel right.


Then the original equation should be:

\(\displaystyle y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}\)
\(\displaystyle y= e^{\frac{x^3}{3}+ C}\)

\(\displaystyle y = e^\frac{x^3}{3}e^C\)

\(\displaystyle y=Ce^{\frac{x^3}{3}}\)

\(\displaystyle 1=Ce^{\frac{1}{3}}\)

\(\displaystyle \frac{1}{e^{\frac{1}{3}}} = C\)
 
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Apr 2010
18
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How did you get from
\(\displaystyle y=e^\frac{x^3}{3} e^C\)

to
\(\displaystyle y=Ce^\frac{x^3}{3}\)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
\(\displaystyle e\) to the power of an arbitrary constant remains an arbitrary constant. Therefore we can say \(\displaystyle e^C = C\)
 
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Apr 2010
18
3
Ahh, I understand now, thank you to all.
 
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