1
$\begingroup$

I needed help with this Differential Equation, below:

$dy/dt = t + y, \text{ with } y(0) = -1$

I tried $dy/(t+y) = dt$ and integrated both sides, but it looks like the $u$-substitution does not work out.

  • 0
    I don't know about you, but for me it's easier to remember a method for solving a problem than a formula.2012-03-18

2 Answers 2

4

This equation is not separable. In other words, you can't write it as $f(y)\;dy=g(t)\;dt$. A differential equation like this can be solved by integrating factors. First, rewrite the equation as:

$\frac{dy}{dt}-y=t$

Now we multiply the equation by an integrating factor so we can use the product rule, $d(uv)=udv+vdu.$ For this problem, that integrating factor would be $e^{-t}$.

$e^{-t}\frac{dy}{dt}-e^{-t}y=\frac d{dt}(e^{-t}y)=te^{-t}$

$e^{-t}y=\int te^{-t}dt=-te^{-t}+\int e^{-t}dt=-te^{-t}-e^{-t}+C$

$y=Ce^t-t-1$

For this specific problem, we could also follow Iasafro's suggestion.

$z=y+t,\frac{dz}{dt}=\frac{dy}{dt}+1,\frac{dy}{dt}=\frac{dz}{dt}-1$

$\frac{dz}{dt}-1=z,\frac{dz}{dt}=z+1,\frac{dz}{z+1}=dt$

As you can see, this substitution resulted in a separable equation, allowing you to integrate both sides.

2

This is a first order linear differential equation so general solution is given by :

$y=\frac{\int u(t)\cdot t \,dt +C}{u(t)} ~\text{where}~ u(t)=e^{-\int dt}$