Solving a System of Differential Equations

Substitution Method

The Problem

dx/dt = 5x + 2y
dy/dt = 5y + e5t
dz/dt = 3y + 5z

Find the general solution of the system

Substitution Method

This is a triangular system — the second equation depends only on y, so we solve it first, then substitute into the remaining equations.

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Step 1: Solve for y(t)
The second equation is a first-order linear equation in y only

Guiding Question

Solve the equation:

y' − 5y = e5t

Hint: Use the integrating factor e−5t

y(t) = · e5t + · e5t

Enter: t for a function of t, or B for a constant

🧐 Digital Friend - Step 1

Step 1 Solution

The equation:
y' − 5y = e5t
Multiply by integrating factor e−5t:
e−5ty' − 5e−5ty = 1
The left side is the derivative of a product:
(e−5ty)' = 1
Integration:
e−5ty = t + B

The Solution

y(t) = te5t + Be5t
Step 2: Solve for x(t)
Substitute y(t) into the first equation and solve for x

Guiding Question

Substitute y(t) = te5t + Be5t into the equation:

x' − 5x = 2te5t + 2Be5t

Hint: Again use the integrating factor e−5t

x(t) = · e5t + · e5t + · e5t

Enter: t^2 for t², 2Bt for 2Bt, A for a constant

🧐 Digital Friend - Step 2

Step 2 Solution

The equation after substitution:
x' − 5x = 2te5t + 2Be5t
Multiply by integrating factor e−5t:
(e−5tx)' = 2t + 2B
Integration:
e−5tx = t² + 2Bt + A

The Solution

x(t) = t²e5t + 2Bte5t + Ae5t
Step 3: Solve for z(t)
Substitute y(t) into the third equation and solve for z

Guiding Question

Substitute y(t) = te5t + Be5t into the equation:

z' − 5z = 3te5t + 3Be5t
z(t) = · e5t + · e5t + · e5t
🧐 Digital Friend - Step 3

Step 3 Solution

The equation after substitution:
z' − 5z = 3te5t + 3Be5t
Multiply by integrating factor e−5t:
(e−5tz)' = 3t + 3B
Integration:
e−5tz = (3/2)t² + 3Bt + C

The Solution

z(t) = (3/2)t²e5t + 3Bte5t + Ce5t