dydx6y and find the particular solution satisfying the initi
dy/dx=.6y
and find the particular solution satisfying the inition condition
y(0)=4
y(x)=?
and find the particular solution satisfying the inition condition
y(0)=4
y(x)=?
Solution
dy/dx = 0.6y
Hence dy/y = 0.6dx
Hence if we integrate both sides we get
ln y = 0.6x + c
Hence y = Ke0.6x
Now as per given condition y = 4 when x = 0;
This gives 4 = K*e0
Hence k = 4
We have y(x) = 4e0.6x
