Derive ODE for $\phi (u)$ Given a system which is described by 
$$10 \phi'(t)+\phi(t)=\phi_H(t)$$
$$\phi_H'(t)+0.5\phi_H(t)=0.5i(t)$$
$$i(t) = 0.5u^2(t).$$
Derive a ODE for $\phi(u(t)).$
My approch is du just put $u(t)$ into the system, but this leads to nothing. Hints would be very nice.
Greetings.
 A: You can solve it by taking the derivative of the first equation and substitute all the other equation into it until you are only left with derivatives of $\phi$ and $u$.
And if your last equation would have $i'$ instead of $i$ on the left hand side then you would have to take the derivative one extra time.
A: You have the following equations:
$$10 \phi'(t)+\phi(t)=\phi_H(t)$$
$$\phi_H'(t)+0.5\phi_H(t)=0.5i(t)$$
$$i(t) = 0.5u^2(t)$$
First, determine the derivative of the first equation.
$$10 \phi''(t)+\phi'(t)=\phi_H'(t)$$
Add this equation and $0.5$ times your first equation:
$$[10 \phi''(t)+\phi'(t)]+0.5[10 \phi'(t)+\phi(t)]=\phi_H'(t)+0.5\phi_H(t).$$
Note, that the right-hand side is the second of your equations, hence the right-hand side is $0.5i(t)=0.5(0.5u^2(t))=0.25u^2(t)$. So we obtain:
$$[10 \phi''(t)+\phi'(t)]+0.5[10 \phi'(t)+\phi(t)]=0.25u^2(t).$$
$$\implies 10 \phi''(t)+6\phi'(t)+0.5\phi(t)=0.25u^2(t)$$
$$\implies \phi''(t)+0.600\phi'(t)+0.050\phi(t)=0.025u^2(t)$$

An alternative approach is using transfer functions. The Laplace transform (zero initial conditions) of your equations is given by:
$$(10s+1)\phi(s)=\phi_H(s) \implies F_1(s)=\phi(s)/\phi_H(s)=1/(10s+1)$$
$$(s+0.5)\phi_H(s)=0.5i(s) \implies F_2(s)=\phi_H(s)/i(s)=0.5/(s+0.5)$$
$$i(s) = 0.5u^2(s) \implies F_3(s)=i(s)/u^2(s)=0.5$$
Note, that $F_1F_2F_3=\frac{\phi(s)}{u^2(s)}=\frac{0.5^2}{(10s+1)(s+0.5)}$. From this it is easy to get the differential equation for $\phi(t)$, with $u^2(t)$ as input.
Remark: I am using $u^2(s)$ as a symbolic representation for the Laplace transform of $u^2(t)$.
