Eliminating a variable from system of ODE's

[gabriels-horn]

Homework Statement

The writing below the equation is the correct order of the constants directly above it.

The Attempt at a Solution

<br /> h=\frac{-g\prime-a_1g+E(t)}{a_2}<br />

So I solved for h in the dg/dt equation, and plug this into the h in the dh/dt equation. My question is where do I go from here to satisfy the relation given in the problem above.

 

[vela]

Homework Statement

View attachment 25711 The writing below the equation is the correct order of the constants directly above it.

The Attempt at a Solution

<br /> h=\frac{-g\prime-a_1g+E(t)}{a_2}<br />

So I solved for h in the dg/dt equation, and plug this into the h in the dh/dt equation. My question is where do I go from here to satisfy the relation given in the problem above.

Now you want to differentiate your expression for h and plug that result into the LHS of the dh/dt equation to get everything in terms of g and its derivatives.

 

[gabriels-horn]

Now you want to differentiate your expression for h and plug that result into the LHS of the dh/dt equation to get everything in terms of g and its derivatives.

So <br /> h=\frac{-g\prime-a_1g+E(t)}{a_2}<br />

becomes

<br /> h\prime=-g\prime\prime-g\prime+E\prime(t)<br />

which replaces dh/dt in the original equation?

 

[vela]

You made a few mistakes. What happened to a₁ and a₂?

 

[gabriels-horn]

You made a few mistakes. What happened to a₁ and a₂?

sorry, still trying to get used to latex

h\prime=\frac{-g\prime\prime-a_1g\prime+E\prime(t)}{a_2}

So this result replaces the left hand side of the dh/dt equation in the original system.

Also, dg/dt from the original equation dg/dt=-a_1g-a_2h+E(t)
becomes

g\prime\prime=-a_1g\prime-a_2h\prime+E\prime

and plug in dh/dt into this equation, making

g\prime\prime=-a_1g\prime-a_2(-a_3g+a_4h)+E\prime