First order differential equation

[Felipe Lincoln]

Homework Statement

Solve the following differential equation such that $x(0)=1$.
$\dfrac{dx}{dt} + 2tx = 3e^{-t^2}+t$

Homework Equations

Integrating factor:
$\mu(t) = exp\left(\int_0^t2t \right)$

The Attempt at a Solution

I used the integrating factor and then got the solution $x(t) = 3te^{-t^2}+\dfrac{1}{2} + C$ and using the initial condition I got $x(t) = 3te^{-t^2}+1$ but if I replace this result into the differential equation I get 2t = t. I first tried to solve it again and got the same solution. Maybe the initial condition should be 1/2 ?

 

[Dick]

Homework Statement

Solve the following differential equation such that $x(0)=1$.
$\dfrac{dx}{dt} + 2tx = 3e^{-t^2}+t$

Homework Equations

Integrating factor:
$\mu(t) = exp\left(\int_0^t2t \right)$

The Attempt at a Solution

I used the integrating factor and then got the solution $x(t) = 3te^{-t^2}+\dfrac{1}{2} + C$ and using the initial condition I got $x(t) = 3te^{-t^2}+1$ but if I replace this result into the differential equation I get 2t = t. I first tried to solve it again and got the same solution. Maybe the initial condition should be 1/2 ?

You'd better show us how you solved that because I don't get a simple additive constant $C$. The $C$ should be multiplied by something related to the integrating factor.

 

[LCKurtz]

I agree with Dick. Check that you treated the $C$ correctly at the step where you solved for $x(t)$.

 

[Felipe Lincoln]

I agree with both of you, my $C$ isn't correct, but I do not think it would matter, if my solution was $x(t) = 3te^{-t^2}+ C$ I would get the same result applying the initial conditions.

 

[mpresic3]

you made an algebraic mistake when you had to integrate exp ( t^2) ( 3 exp(-t^2) ) i.e integrate 3 dt to get 3t now do rest of the calculation

 

[LCKurtz]

I agree with both of you, my $C$ isn't correct, but I do not think it would matter, if my solution was $x(t) = 3te^{-t^2}+ C$ I would get the same result applying the initial conditions.

Actually, your $C$ is correct, but your solution is not. Your difficulty has nothing to do with the initial condition. Post your solution steps here and someone will show your mistake.

 

[SammyS]

Homework Statement

Solve the following differential equation such that $x(0)=1$.
$\dfrac{dx}{dt} + 2tx = 3e^{-t^2}+t$

Homework Equations

Integrating factor:
$\mu(t) = exp\left(\int_0^t2t \right)$

The Attempt at a Solution

I used the integrating factor and then got the solution $x(t) = 3te^{-t^2}+\dfrac{1}{2} + C$ and using the initial condition I got $x(t) = 3te^{-t^2}+1$ but if I replace this result into the differential equation I get 2t = t. I first tried to solve it again and got the same solution. Maybe the initial condition should be 1/2 ?

It looks like you obtained the correct integrating factor $\displaystyle \ \mu (t) =e^{t^2}$, and did the integration correctly. At that point, you should have included a constant of integration, $\ C\,.\$ Also, at that point, you could have applied the boundary condition to evaluate $\ C\,.\$

Alternatively, rather than evaluating $\ C\$ at that point, it seems that you then solved for $\ x(t)\$ by multiplying by $\displaystyle \ e^{-t^2} \ ($ or dividing by $\displaystyle \ e^{t^2}\,).\$ It seems that in doing this you forgot to also multiply (or divide) $\ C\$ by this factor.