# finding solution by initial conditions

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• Aug 18th 2009, 10:05 AM
dat1611
finding solution by initial conditions
The solution of a certain differential equation is of the form

y(t)=aexp(4t)+bexp(6t), where a and b are constants.
The solution has initial conditions y(0)=1 and y'(0)=3.
Find the solution by using the initial conditions to get linear equations for a and b.

can anyone help me with this one.
• Aug 18th 2009, 10:17 AM
Random Variable
$\displaystyle y(t) = Ae^{4t} + Be^{6t}$

$\displaystyle y(0) = 1= Ae^{0} + Be^{0} = A+B$

$\displaystyle y'(t) = 4Ae^{4t} + 6Be^{6t}$

$\displaystyle y'(0) =3 = 4Ae^{0} + 6Be^{0} = 4A+6B$

so solve the system

A+B=1
4A+6B=3
• Aug 18th 2009, 10:31 AM
dat1611
i dont know really what to do.
• Aug 18th 2009, 10:36 AM
Random Variable
multipy the first equation by -4 and add it to the second equation to get $\displaystyle 2B=-1$ or $\displaystyle B= -\frac{1}{2}$

then $\displaystyle A - \frac{1}{2} = 1$ or $\displaystyle A = \frac{3}{2}$

so $\displaystyle y(t) = \frac{3}{2}e^{4t} - \frac{1}{2}e^{6t}$