Question

Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dxdt+2x−2=0

Answer 1

Given the differential equation


`5t^3 (dx)/(dt) +2x-2=0`

The solution is as follows:


`5t^3 (dx)/(dt) +2x-2=0 \\ \\ \Rightarrow5t^3 (dx)/(dt) =2-2x \\ \\ \Rightarrow (5)/(2-2x) dx= (1)/(t^3) dt \\ \\ \Rightarrow \int {(5)/(2-2x) } \, dx = \int {(1)/(t^3)} \, dt \\ \\ \Rightarrow- (5)/(2) \ln(2-2x)=- (1)/(2t^2) +A \\ \\ \Rightarrow\ln(2-2x)= (1)/(5t^2) +B\\ \\ \Rightarrow2-2x=Ce^{(1)/(5t^2)} \\ \\ \Rightarrow 2x=Ce^{(1)/(5t^2)}+2 \\ \\ \Rightarrow x=De^{(1)/(5t^2)}+1`