Linear First-Order Differential Equations
A linear first-order ode has the form:
where g(t) and h(t) are given functions.
Solution Procedure
Rewrite the problem in the form:
We multiply the above equation by an integrating factor:
The derivative of u(t), which we will use below, is
(Recall the derivative of an exponential is the exponential multiplied by the derivative of its argument. The argument is an integral, which we differentiate by the Fundamental Theorem of Calculus.) The new equation is
Why did we do this? Look at the left-hand side of the equation. Let z(t)=u(t)y(t). Using the product rule and the result above for u'(t), we have
Hence, equation (*) becomes
This is now in the form of a directly integrable equation since both u(t) and h(t) are known. Applying the indefinite form of the Fundamental Theorem of Calculus, we obtain
Solving for y(t)=z(t)/u(t), we have
We can also write the answer in the form of a definite integral. Applying the Fundamental Theorem of Calculus we have
Solving for y(t), we have
Example
ODE:
The integrating factor is
The indefinite integral form of the solution is
The integral is
Hence, we have
Applying the initial condition y(0)=1, we find that C=3/2.
Alternatively, we can use the definite integral form of the solution. We have
Using the fact that u(0)=1 and y(0)=1, we have
The integral is
Hence, we obtain
