Differential Equations Exercise 27

We want to solve the differential equation

$$ y'' - y' = 3 $$

This is a non-homogeneous second-order differential equation.

There are two standard approaches to solving it.

Method 1

The solution can be found by combining the general solution of the homogeneous equation with a particular solution of the non-homogeneous one.

1] The homogeneous solution

The associated homogeneous equation is:

$$ y'' - y' = 0 $$

The corresponding characteristic equation is:

$$ z^2 - z = 0 $$

Factoring, we obtain:

$$ z(z-1) = 0 $$

Thus, the roots are $z_1=0$ and $z_2=1$.

Therefore, the homogeneous solution is:

$$ y_h = c_1 e^{0 \cdot x} + c_2 e^{1 \cdot x} $$

which simplifies to

$$ y_h = c_1 + c_2 e^x $$

2] The particular solution

To find a particular solution, we use the method of undetermined coefficients.

The forcing term is the constant $3$. Since the equation does not involve $y$ itself, a suitable trial solution is of the form:

$$ y_p = Ax $$

Substituting into the differential equation:

$$ y_p'' - y_p' = 3 $$

gives $y_p' = A$ and $y_p'' = 0$, so

$$ 0 - A = 3 $$

which yields $A = -3$.

Hence, the particular solution is:

$$ y_p = -3x $$

3] The general solution

The general solution of the original equation is the sum of the homogeneous and particular solutions:

$$ y = y_h + y_p $$

Substituting $y_h = c_1 + c_2 e^x$ and $y_p = -3x$, we obtain

$$ y = c_1 + c_2 e^x - 3x $$

This is the general solution of the differential equation.

Method 2

The given equation is a second-order equation that does not explicitly involve $y=f(x)$.

$$ y'' - y' = 3 $$

We reduce the order by introducing the auxiliary variable $u = y'$:

$$ y'' - u = 3 $$

Since $u = y'$, it follows that $u' = y''$, so the equation becomes:

$$ u' - u = 3 $$

This is now a first-order linear differential equation.

It has the standard form $u' - A(x)u = B(x)$ with $A(x)=1$, and can be solved using the integrating factor method.

The integrating factor is

$$ M(x) = e^{\int -1 \, dx } = e^{-x} $$

Multiplying through by $M(x)$ gives:

$$ e^{-x}(u' - u) = 3 e^{-x} $$

which can be rewritten as:

$$ D\!\left(\frac{u}{e^x}\right) = \frac{3}{e^x} $$

Integrating both sides:

$$ \frac{u}{e^x} = c_1 - \frac{3}{e^x} $$

Multiplying through by $e^x$ yields:

$$ u = c_1 e^x - 3 $$

Since $u=y'$, we have:

$$ y' = c_1 e^x - 3 $$

Integrating once more gives:

$$ y = c_2 + c_1 e^x - 3x $$

which is exactly the same general solution obtained with Method 1.