Linear First-Order Differential Equations

A linear first-order ode has the form:

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where g(t) and h(t) are given functions. In addition to the solution procedure outlined below, it may also be possible to solve the problem using the Laplace Transform technique.

Solution Procedure

Rewrite the problem in the form:

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We multiply the above equation by an integrating factor:

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The derivative of u(t), which we will use below, is

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(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

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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

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Hence, equation (*) becomes

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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

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Solving for y(t)=z(t)/u(t), we have

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We can also write the answer in the form of a definite integral. Applying
the Fundamental Theorem of Calculus we have

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Solving for y(t), we have

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Example

ODE:

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The integrating factor is

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The indefinite integral form of the solution is

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The integral is

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Hence, we have

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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

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Using the fact that u(0)=1 and y(0)=1, we have

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The integral is

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Hence, we obtain

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