Question

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

Answer 1

Answer:

$y=Ce^{-x}$

Step-by-step 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=$\int e^{3x^2dx}=e^{\frac{3x^3}{3}=e^x$

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

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

$y=Ce^{-x}$