Hey guys, I need WolframAlpha to solve:

x(t)=(C_1)e^(-2.74031*t)+(C_2)e^(-1.45969*t)

when x(0)=2, and x'(0)=1.

How can I do this?

## How To

### Solving equation for 2 variables with constraints?

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**1**of**1**• [ 2 posts ]### Solving equation for 2 variables with constraints?

by » Mon Feb 27, 2012 3:12 am- Norm850
**Posts:**1**Joined:**Mon Feb 27, 2012 3:10 am

### Re: Solving equation for 2 variables with constraints?

by » Tue Feb 28, 2012 3:19 amA bit messy but I did this:

W|A: c1*e^(-2.74031*0)+c2*e^(-1.45969*0)==2

Results included: c1+c2 = 2

And W|A: d(c1*e^(-2.74031*t)+c2*e^(-1.45969*t))/dt

Gives: -2.74031 c1 e^(-2.74031 t)-1.45969 c2

W|A: -2.74031*c1*e^(-2.74031*0)-1.45969*c2==1

Yields (with c1 taken as a variable): -2.74031 c1-1.45969 c2 = 1

Finally, W|A: {c1+c2 = 2, -2.74031 c1-1.45969 c2 = 1}

Results include the solution: c1~~-3.06053, c2~~5.06053

I substituted c1, c2 for C_1, C_2 as W|A can get confused with other subjects sometimes, chemistry for instance.

Ideally W|A would do all this if you entered just:

{f(t)=c1*e^(-2.74031*t)+c2*e^(-1.45969*t),f(0)=2,f'(0)=1}

But, in this case, no helpful result is obtained just posing the above direct query.

W|A: c1*e^(-2.74031*0)+c2*e^(-1.45969*0)==2

Results included: c1+c2 = 2

And W|A: d(c1*e^(-2.74031*t)+c2*e^(-1.45969*t))/dt

Gives: -2.74031 c1 e^(-2.74031 t)-1.45969 c2

W|A: -2.74031*c1*e^(-2.74031*0)-1.45969*c2==1

Yields (with c1 taken as a variable): -2.74031 c1-1.45969 c2 = 1

Finally, W|A: {c1+c2 = 2, -2.74031 c1-1.45969 c2 = 1}

Results include the solution: c1~~-3.06053, c2~~5.06053

I substituted c1, c2 for C_1, C_2 as W|A can get confused with other subjects sometimes, chemistry for instance.

Ideally W|A would do all this if you entered just:

{f(t)=c1*e^(-2.74031*t)+c2*e^(-1.45969*t),f(0)=2,f'(0)=1}

But, in this case, no helpful result is obtained just posing the above direct query.

- jyellott
**Posts:**135**Joined:**Mon Aug 17, 2009 2:35 pm

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