Question

Find the general solution of the differential equation: y' + 3x^2 y = 0

Answer 1

Answer:

The general solution of the differential equation y' + 3x^2 y = 0 is:

$y=e^{-x^3+C}$

Step-by-step explanation:

This equation its a Separable First Order Differential Equation, this means that you can express the equation in the following way:

$\frac{dy}{dx} = f_1(x)*f_2(x)$, notice that the notation for y' is changed to $\frac{dy}{dx}$

Then you can separate the equation and put the x part of the equation on one side and the y part on the other, like this:

$\frac{1}{f_2(x)}dy=f_1(x)dx$

The Next step is to integrate both sides of the equation separately and then simplify the equation.

For the differential equation in question y' + 3x^2 y = 0 the process is:

Step 1: Separate the x part and the y part

$\frac{1}{y}dy=3x^2}dx$

Step 2: Integrate both sides

$\int\frac{1}{y}dy=\int 3x^2}dx$

Step 3: Solve the integrals

$Ln(y)+C=-x^3+C$

Simplify the equation:

$Ln(y)=-x^3+C$

To solve the Logarithmic expression you have to use the exponential e

$e^{Ln(y)}=e^{-x^3+C}$

Then the solution is:

$y= e^{-x^3+C}$