I have the differential equation:
$$\frac{dN_b}{dt}=\lambda_aNa(0)exp(-\lambda_at)-\lambda_bN_b(1)$$
I am given that the solution to this is:
$$N_b(t)=\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[exp(-\lambda_at)-exp(-\lambda_bt)](2)$$
I am told that differentiating (2) against time, and then substituting into (1) will return the same answer. However, no matter what I do I cannot figure out how this works. I do not understand what I am doing wrong. I know the derivative of (2) is:
$$\frac{dN_b}{dt}=\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[-\lambda_aexp(-\lambda_at)+\lambda_bexp(-\lambda_bt)](3)$$
Substituting (3) into (1) and (2) into (1) provides:
$$\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[-\lambda_aexp(-\lambda_at)+\lambda_bexp(-\lambda_bt)]=\lambda_aNa(0)exp(-\lambda_at)-\lambda_b\frac{\lambda_a}{\lambda_b-\lambda_a}\cdot Na(0)[exp(-\lambda_at)-exp(-\lambda_bt)]$$
No matter how much I simplify, I never get out $$0=0$$
This is a well-known and well-proven differential equation relating to radioactive decay. Equation 2 has been shown to be a solution to Equation 1, but my professor wants us to solve it through differential form instead of solving the ordinary differential equation. Can someone please help me? Thank you so much.
Starting point: $$ \dot{N}_b = \lambda_a N_{a,0} e^{-\lambda_a t} - \lambda_b N_b $$ $$ N_b = \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) $$ Take a derivative of the second equation: $$\dot{N}_b = \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a}\Big(-\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big).$$ Plug the latter two equations into the first equation: $$ \frac{\lambda_a N_{a,0}}{\lambda_b-\lambda_a} \Big( -\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big) = \lambda_a N_{a,0}e^{-\lambda_a t} - \frac{\lambda_a \lambda_b N_{a,0}}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) .$$ Cancel common factors: $$ \frac{1}{\lambda_b-\lambda_a} \Big( -\lambda_a e^{-\lambda_a t} + \lambda_b e^{-\lambda_b t}\Big) = e^{-\lambda_a t} - \frac{\lambda_b}{\lambda_b-\lambda_a}\Big( e^{- \lambda_a t} - e^{-\lambda_b t}\Big) .$$ Collect terms involving linearly independent factors $$ \Big(\frac{-\lambda_a}{\lambda_b-\lambda_a} -1 + \frac{\lambda_b}{\lambda_b-\lambda_a}\Big)e^{-\lambda_a t} = \Big(\frac{\lambda_b}{\lambda_b-\lambda_a}-\frac{\lambda_b}{\lambda_b-\lambda_a}\Big)e^{-\lambda_b t}$$ Simplify: $$ 0 = 0$$