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Problem

I need to solve differential equation which is defined as:

$$ y'=2+y,\quad y(0)=2 $$

Attempt to solve

Now we could write the equation in form of:

$$ \frac{dy(x)}{dx}=y(x)+2$$ Divide by $y(x)+2$ $$ \frac{\frac{dy(x)}{dx}}{y(x)+2}=1 $$ Now integral both sides $$ \int{\frac{\frac{dy(x)}{dx}}{y(x)+2}}dx = \int{1} dx $$ Now here is the part i don't understand. This should equal the following: $$ \int{\frac{1}{y+2}}dy=\int{1}dx $$ Solving left side with u-substitution shouldn't be problematic but the problem is i don't have good understanding what's going on here ? How do you get

$$ \int{\frac{\frac{dy(x)}{dx}}{y(x)+2}}dx=\int{\frac{1}{y+2}}dy $$

Another thing i don't understand is the notation used in the problem. there is $y(0)=2$ what i am suppose to do with this information ?

The solution set for this equation is:

enter image description here

But again i don't think i can fully understand this.

If someone could provide some explanation whats going on here that would be highly appreciated.

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  • $\begingroup$ Sometimes, it is taken for granted that $y(x)$ can be written as $y$. Hence, $$\frac{dy(x)}{dx} = y(x)+2 \equiv \frac{dy}{dx} = y+2$$ which can be easily solved. $\endgroup$
    – user371838
    Dec 4 '17 at 12:17
  • $\begingroup$ Maybe of interest: math.stackexchange.com/questions/27425/… $\endgroup$ Dec 4 '17 at 13:25
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As written the symbol $y$ is doing double duty: as the indeterminate function in the original differential equation, and as the variable of substitution. To untangle what's going on here, it might help to introduce a new symbol for the substitution variable. Your integral is $$\int \frac{\frac{d}{dx} y(x)}{y(x) + 2} dx .$$ Substituting $$u = y(x) , \qquad du = \frac{d}{dx} y(x) \, dx ,$$ (here, as usual, we have computed $du$ using the chain rule) gives that the integral is $$\int \frac{du}{u + 2} = \log\left\vert u + 2 \right\vert + C' = \log\left\vert y(x) + 2 \right\vert + C'.$$


Separately, note that we can change simplify slightly our original equation before separating variables, by writing $v(x) = y(x) + 2$. Then, $v'(x) = y'(x)$ and we can write the original equation as $$v'(x) = v(x) .$$

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$$\int{\frac{\frac{dy(x)}{dx}}{y(x)+2}}dx=\int{\frac{1}{y(x)+2}\cdot \frac{dy(x)}{dx}}dx=\int\frac{1}{y(x)+2}dy(x)$$

$dx$ in the numerator simplifies with $dx$ in the denominator

which can be written, for brevity, $$\int{\frac{1}{y+2}}dy$$ but the meaning is the same.

$$\int{\frac{1}{y+2}}dy=\int\,dx$$

$$\log(y+2)=x+C$$

Hope this helps

$$y+2=e^{x+C}$$ to determinate $C$ you substitute $x=0$ and get

$2+2=e^C\to C=\log 4$

so the solution which solves the differential equation is

$$y=4e^x-2$$

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You can solve this equation with an integration factor.

$$y'(x)=2+y(x)$$ $$e^{-x}y'(x)=e^{-x}(2+y(x)) $$ Now write one side as a derivative: $$\Rightarrow (y(x)e^{-x})'=2e^{-x}$$ Now we just integrate, I renamed the variable to $\alpha$ to clarify its just a dummy variable. $x_0$ is just a constant. $$\Rightarrow \int_{x_0}^x (y(\alpha)e^{-\alpha})'d\alpha=\int_{x_0}^x2e^{-\alpha}d\alpha$$ Integrate the expression: $$y(x)e^{-x}-y(x_0)e^{-x_0}=-2e^{-x}+2e^{-x_0}.$$ This solution is not unique, it still has a constant that can have different values. Luckily we have given an initial condition $y(0)=2$. So, when we use it we get a unique solution.

Just put $x_0=0$ and simplify the expression above: $$y(x)e^{-x}-2=-2e^{-x}+2 \Rightarrow y(x)=-2+4e^x.$$

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  • $\begingroup$ While this is a decent solution, it does not address any of OP's questions. $\endgroup$
    – Dylan
    Dec 4 '17 at 20:09

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