Question

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

Answer 1

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

`y=e^(-x^3+C)`

Explanation:

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

`(dy)/(dx) = f_1(x)*f_2(x)`, notice that the notation for y' is changed to
`(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:

`(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

`(1)/(y)dy=3x^2}dx`

Step 2: Integrate both sides

`\int(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)`