Question

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

Answer 1

`y=Ce^(-x)`

Explanation:

We are given that a differential equation

`y'+3x^2y=0`

We have to find the general solution of given differential equation

General differential equation of first order and first degree is given by

`y'+p(x)y=Q(x)`

Compare given differential equation with the general equation

Then , we get
`P(x)=3x^2`,Q(x)=0

Integration factor=
`\int e^(3x^2dx)=e^{(3x^3)/(3)=e^x`

`y\cdot I.F=\int Q(x)\cdot I.F+C`

Substitute the values then we get

`y\cdot e^x=0+C`

`ye^x=C`

`y=Ce^(-x)`

Hence, the general solution of given differential equation

`y=Ce^(-x)`