general solutions to differential equations of first order

Question

general solution to differential equations:

$$\frac{dy}{dx}+e^{2x}y^2=0$$

$$\frac{dy}{dx}= -e^{2x}y^2$$

$$\int \frac{1}{y^2} dy= \int -e^{2x} dx+c$$

$$-\frac{1}{y}= -0.5\ e^{2x}+c$$

$$y= 2e^{-2x}+c $$

Is this correct?

Answer 1

Be careful,

$$\frac{dy}{dx}+e^{2x}y^2=0$$

$$ \Longleftrightarrow \frac{dy}{dx}= -e^{2x}y^2$$

$$\Longleftrightarrow \int \frac{1}{y^2} dy= \int -e^{2x} dx$$

Why do you put a constant c in your integration? The constant comes by integrating, I mean after the integration is completed, not before.

$$\Longleftrightarrow -\frac{1}{y}= -0.5\ e^{2x}+c_{0}$$

$$\Longleftrightarrow \frac{1}{y}= 0.5\ e^{2x}+c_{1}$$

with $$ c_{1} = -c_{0} $$

Finally,

$$\Longleftrightarrow {y}= \frac{2}{e^{2x}+c_{1}}$$