Answer:
The general solution of the differential equation y' + 3x^2 y = 0 is:
[tex]y=e^{-x^3+C}[/tex]
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:
[tex]\frac{dy}{dx} = f_1(x)*f_2(x)[/tex], notice that the notation for y' is changed to [tex]\frac{dy}{dx}[/tex]
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:
[tex]\frac{1}{f_2(x)}dy=f_1(x)dx[/tex]
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
[tex]\frac{1}{y}dy=3x^2}dx[/tex]
Step 2: Integrate both sides
[tex]\int\frac{1}{y}dy=\int 3x^2}dx[/tex]
Step 3: Solve the integrals
[tex]Ln(y)+C=-x^3+C[/tex]
Simplify the equation:
[tex]Ln(y)=-x^3+C[/tex]
To solve the Logarithmic expression you have to use the exponential e
[tex]e^{Ln(y)}=e^{-x^3+C}[/tex]
Then the solution is:
[tex]y= e^{-x^3+C}[/tex]
