In the previous two lessons, we learned that Laplace transforms can be an effective tool when solving certain differential equations involving discontinuous functions and functions defined on split domains. In this, the final Differential Equation lesson, we will apply Laplace transforms to solving systems of simultaneous differential equations and employ this method to a circuit problem. There are other methods for solving systems of equations - see chapter 8 on matrix methods - but Laplace transforms are far more versatile being capable of solving systems involving discontinuities, necessary for common circuit applications.

The general procedure is to find the Laplace transform of each differential equation, set up a system of equations in terms of the transforms, solve for each and find the inverse transform for the solution. An example is in order.

Solve the homogeneous differential system : x'( t ) = -3x( t ) + y( t ) & y'( t ) = 2x( t ) - 4y( t ), x( 0 ) = 0 ,y( 0 ) = 1.

Now let's do a system of nonhomogeneous differential equations. Please note that the 4e^2t and 4e^4t functions of t make the differential equations nonhomogeneous.

We will now apply this technique to solving an LR circuit with two inductors and two resistors. Here is the circuit diagram with a source at E volts, R₁ & R₂ ohms, and L₁ & L₂ henries.

If E = e( t ) = 100sin( t ) and the other values are constants as given, then the solution to the circuit for i1( t ) and i2( t ) is as follows:

Here is the graph of the two currents.

 

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Things to know

• Solve a system of differential equations with Laplace transforms.
• Solve an LR circuit or RC circuit two nonhomogeneous differential equations.
• Solve the following system: x'( t ) = x + y, y'( t ) = x - y, x( 0 ) = 1, y( 0 ) = 1.
• Solve the above LR circuit if e( t ) = 12.

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