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I am studying differential equations from a book called Differential Equations (Schaum's outlines). In it, it says if we have a separable differential equation:

$$ A(x) dx +B(y) dy=0, \ y(x_0)=y_0 $$

then, the solution to the initial-value problem can be obtained by $$ \int_{x_0}^{x}A(x)dx+ \int_{y_0}^{y} B(y)dy=0 $$

And that's my problem. I was expecting something like $$ \int_{x_0}^{x}A(x)dx+ \int_{x_0}^{x} B(y)dy=0 $$ as we integrate the first equation from $ x_o $ to $ x $.

I admit that what i am stating is not so intuitive (in fact, the second equation makes more sense than the third). But how can i prove that $$ \int_{x_0}^{x} B(y(x)) \ dy(x) = \int_{y_0}^{y} B(y)dy $$

2 Answers2

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I think it's easier to think about if we rewrite the differential equation as:

$$A(x) +B(y) \frac{dy}{dx}=0$$

Now integrate both sides with respect to $x$ (with limits $x_0$ and $x$) to give:

$$\int_{x_0}^{x}A(x)dx+ \int_{x_0}^{x} B(y(x))\frac{dy}{dx}dx=0$$

In the second integral, make the change of variable $y=y(x)$ (i.e. integration by substitution) and apply your initial condition, so that:

$$ \begin{align} \int_{x_0}^{x} B(y(x))\frac{dy}{dx}dx &=\int_{y(x_0)}^{y(x)} B(y)dy \\ &=\int_{y_0}^{y} B(y)dy \end{align}$$

... as stated!

In your final line, the expression

$$\int_{x_0}^{x} B(y(x)) \ dy(x)$$

doesn't really make sense. The symbols "$dx$" and "$dy$" have a number of different interpretations. Both of the statements $dy=\frac{dy}{dx}dx$ and $dy=y'(x)dx$ make sense, but $dy(x)$ is dodgy.

Malkin
  • 2,113
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Taking the derivative on $x$ of

$$\int_{x_0}^xA(x)\,dx+\int_{x_0}^xB(y)\,dy=C$$ you obtain

$$A(x)+B(x)=0,$$ which is certainly not the original ODE.

On the opposite

$$\int_{x_0}^xA(x)\,dx+\int_{y_0}^{y(x)}B(y)\,dy$$ gives

$$A(x)+B(y(x))\frac{dy(x)}{dx}.$$