First order differential equation

Gold Member

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 ?

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Dick
Homework Helper

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

Gold Member
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.

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
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member

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.