Help with this intro to Differential EQ problem?

Question

Curious if someone could help me with this problem.  My book doesnt really cover it and its an "into" to Diff EQ from my Calc II class.

du/dx=2u+6, u(1)=6

Can I say:
du/2u+6=dx

Then integrate both sides?

ln|2u+6|=x

then

2u+6=e^x

then

u=C(e^x)-3

Then plug 1 in for x and solve for C?

Help anyone?

hfshaw · Accepted Answer

Yes (almost).  You were sloppy with your constant of integration, and you integrated wrong..

You have a "separable" differential equation, meaning a differential equation for which you can separate the dependent and independent variables on either side of the equals sign.

du/dx = 2u + 6, with u(1) = 6

Separate the variables:

du/(u + 3) = 2 dx

Integrate:

ln(u + 3) = 2x + c

where c is the combined constant of integration from both integrals.

u(x) = exp(2x + c) - 3

Remember that exp(a + b) = exp(a)*exp(b), so:

u(x) = C*exp(2x) - 3

where C = exp(c) is just another way of writing the constant of integration.

Now use the initial condition to solve for C:

u(1) = 6 = C*exp(2) - 3

C = 9*exp(-2)

So the solution to this initial value problem is:

u(x) = 9*exp(-2)*exp(2x) - 3

or, perhaps more concisely:

u(x) = 9*exp(2*(x-1)) - 3

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Help with this intro to Differential EQ problem?

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