# first order differential equation

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.

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Have you try a change of variable $z=y+t$? – Josué Tonelli-Cueto Mar 18 '12 at 11:09
Why bother changing the variable, when there is a direct formula for solving the equation? If you do not want to apply the direct formula, then the steps in the proof of the formula will give a quick result. en.wikipedia.org/wiki/… – Beni Bogosel Mar 18 '12 at 12:12
I don't know about you, but for me it's easier to remember a method for solving a problem than a formula. – Mike Mar 18 '12 at 23:29

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.

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

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

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