# What is i(t) from circuit with differential equation?

When t<0 Components in circuit are Jt = 4 A, R1 = R2 = 2 Ω, R3 = 4 Ω ja L = 10 H

At t=0 switch K will be closed and i need to figure out what iL(t) is when t=4

So i tried to figure out iL(t) with differential equation.
First i combined R1 and R2 to get R12=1Ω Then i transformed power supply from current to voltage Et=Jt*R12=J*1Ω =J
Then i combined R12 and R3 to get Rz=5Ω
Now i can create equation which is $$E_t=L*\frac{di(t)}{dt}+R_ti(t)$$ After adding constant i get
$$J_t=10*\frac{di(t)}{dt}+5i(t)$$ First i found the roots $$10r^t+5r=0 -> r=-\frac{1}{2}$$ So for homogeneous part i get $$y^h=C_1*(-\frac{1}{2})^t$$ And for nonhomogeneous part $$J_t=A$$ $$\frac{dJ_t}{dt}=0 , \frac{d^2J_t}{dt^2}=0$$ So i get $$10*0+5=A->A=5$$ For complete equation i get $$y(t)=C_1*(-\frac{1}{2})^t+5$$ To figure out what is C1 i use t=0 $$y(0)=C_1*(-\frac{1}{2})^0+5=4 ->C_1=-1$$ And now for t=4 i get $$y(4)=-1*(-\frac{1}{2})^4+5=4.5$$ I know this answer isn't correct but i don't know what went wrong
Also sorry if its little bit hard to read this, i'm new to this

• You've solved the differential equation incorrectly. The solution should be an exponential function. The simple method is to analyze the circuit at steady-state before and after the switch closes. In steady-state before t=0, what's the current $i_L$? – Shamtam Sep 18 '17 at 14:40

The proper way to form the homogeneous solution is (not what you have given in the fourth equation): $${i_L}_h=C_1*e^{(-\frac{1}{2})t}$$ (Your differential equation is written for $i_L(t)$, it is better not to change it to $y$ midstream.)
For t>0, without knowing $J$, you cannot solve $i_L(t)$. The only specification of $J$ given is that it is 4. So a forced assumption is that $J = 4$ for all time if you want a solution from what are given.
With $J$ being constant, by inspection, $$i_L(t) = \frac{J_t}{5}$$ is a particular solution to the differential equation. You can plug that into your second equation to confirm.
Now the complete solution becomes: $$i_L(t)=C_1*e^{(-\frac{1}{2})t} + \frac{J_t}{5}$$
Now you need the initial condition of $i_L(0)$ to solve for the constant $C_1$. Given $J$ is constant and assuming steady state at t=0, $L$ behaves as a short circuit at t=0. Therefore, the only elements left in the schematic for figuring out the initial condition are $J,\ R_2,\ R_3$, a simple current divider.