Solving 1st order lin. diff. equations $ (LDE_1) $ with continuous coefficient

Let $ y $ be a function of class $ \mathbb{C}^{1}$ on an interval $ I$.

In this part, we will show the function $ y(x) $ under its simplified form $ y$.

As well, let $ a(x)$ be a continuous function and $ f(x)$ any function.

Let $ (E) $ be a linear differential equation or order $ 1 $ with continuous coefficient, and $ (H) $ its associated homogeneous equation:

$$ \Biggl \{ \begin{align*} y' + a(x) y = f(x) \qquad (E) \\ y' + a(x) y = 0 \qquad (H) \end{align*} $$

Solving the homogeneous equation $ (H) $

$ y_h = 0 $ is solution, but considering $ y_h $ as a non-zero function, the following function $y_h$ will be an homogeneous solution of $ (H) $ :

$$ y_h= Ke^{- A(x) } $$

Solving the general equation $ (E) $

Function $y_s$ will be specific solution of $ (E) $ :

$$ y_s= e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt $$

We will have as a total solution of $ (E) $, the addition of these two solutions:

$$ y_t= y_h + y_s $$

Demonstration

Let $ y $ be a function of class $ \mathbb{C}^{1}$ on an interval $ I$.

In this part, we will show the function $ y(x) $ under its simplified form $ y$.

As well, let $ a(x)$ be a continuous function and $ f(x)$ any function.

Solving a first order linear differential equation $(LDE_1)$ with continuous coefficient consists of finding functions $ y : x \longmapsto y(x)$, which are solutions of an equation $ (E) $ of type:

$$ y' + a(x) y = f(x) \qquad (E) $$

We can first find solutions $ y_h $ to the homogeneous equation $ ( H) $ :

$$ y' + a(x) y = 0 \qquad (H) $$

If $ y_h $ is solution of the homogeneous equation $ ( H) $, and $ y_s $ is specific solution to the equation $ ( E) $, then:

$$ \Biggl \{ \begin{align*} y_s' + a(x) y_s = f(x) \qquad (E) \\ y_h' + a(x) y_h = 0 \qquad (H) \end{align*} $$

By performing the operation $ ( H) + (E) $ :

$$ y_h' + y_s' + a(x) y_h + a(x) y_s = f(x) \qquad (H+E) $$

Thanks to the linearity of the derivative, we notice that $ (y_h + y_s) $ remains solution of $ ( E) $, because:

$$ (y_h + y_s)' + a(x) (y_h + y_s) = f(x) $$

As a result, $ y_t $ will be a total solution of $ (E) $:

$$ y_t= y_h + y_s $$

Solving the homogeneous equation $ (H) $

We call homogeneous equation $ (H) $, the equivalent of the equation $ (E) $ but without any right side member.

$$ y' + a(x) y = 0 \qquad (H) $$

Thus, $ y_h = 0 $ est solution évidente, but let us consider $ y_h $ as a non-zero function.

From $ (H) $, let us try to find a function $ (y_h) $ which corresponds:

$$ y_h' + a(x) y_h = 0 \qquad (H) $$

$$ y_h' = - a(x) y_h $$

$$ \frac{y_h' }{y_h}= - a(x) $$

Now, we recognize the derivative of a composite function $ ln(u) $:

$$ ln(u)' = \frac{u' }{u} $$

So, calculating its antiderivative,

$$ \int^x \frac{y_h'(t) }{y_h(t)} \hspace{0.1em} dt= \int^x - a(t)\hspace{0.1em} dt $$

$$ ln \bigl |y_h \bigr|= - A(x) + C \qquad (with \enspace K \in \mathbb{R}) $$

$$ y_h= e^{- A(x) + C} $$

$$ y_h= Ke^{- A(x) } $$

We then have as an homogeneous solution for $ (H) $:

$$ y_h= Ke^{- A(x) } $$

There are indeed an infinite number of solutions for $ y_h $ , corresponding to ignorance of the value of the constant $K$. he latter can be set subsequently based on the existence of initial conditions.

Verification of the homogeneous solution $ y_h $

If $ y_h $ is solution of $ (H) $, then:

$$ y_h' + a(x) y_h = 0$$

Now, we apply the product of two functions:

$$ y_h' + a(x) y_h = -a (x) K e^{- A(x) } + a(x)Ke^{- A(x) } $$

$$ y_h' + a(x) y_h = 0 $$

$ y_h $ is definitely a homogeneous solution of $ (H) $.

Solving the general equation $ (E) $

Knowing a solution to the homogeneous equation $ (H) $, we can start from this to find specific solutions of $ (E) $.

The method of the variations of parameters

Indeed, when looking for a specific solution $ y_s $ de type :

$$ y_s= K(x) e^{- A(x) } \qquad (y_s) $$

By applying the method of the variations of parameters (or Lagrange's method), we do have:

$$ y_s' + a(x) y_s = K'(x) e^{- A(x) } -a (x) K(x) e^{- A(x) } + a(x)K(x) e^{- A(x) } $$

$$ y_s' + a(x) y_s = K'(x) e^{- A(x) } $$

Given the following system:

$$ \Biggl \{ \begin{align*} y' + a(x) y = f(x) \qquad (E) \\ y_s' + a(x) y_s = K'(x) e^{- A(x) } \qquad (E_p) \end{align*} $$

Thanks to both equations $ (E) $ and $ (E_p) $, we deduce an equation to solve, to determine $ K(x) $ :

$$ K'(x) e^{- A(x) } = f(x) $$

$$ K'(x) = f(x)e^{A(x) } $$

Let us antiderivate each member:

$$ K(x)= \int^x f(t)e^{A(t) } \hspace{0.1em} dt \qquad (K) $$

We inject $ (K) $ into $ (y_s) $:

$$ y_s= e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt $$

We then have as a specific solution for $ (E) $:

$$ y_s= e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt $$

Verification of the specific solution $ y_s $

if $ y_s $ is solution of $ (E) $, then:

$$ y_s' + a(x) y_s = f(x)$$

We again apply the product of two functions:

$$ y_s' + a(x) y_s = -a (x) e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt + f(x)e^{A(x)} e^{- A(x) } + a(x) e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt $$

$$ y_s' + a(x) y_s = \hspace{0.2em} \underbrace{-a (x) e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt + a(x) e^{- A(x) } \int^x f(t)e^{A(t) } \hspace{0.1em} dt } _\text{ $ = \ 0$} \enspace + \enspace \underbrace{f(x)e^{A(x)} e^{- A(x) }} _\text{ $ = \ f(x)$} $$

So,

$$ y_s' + a(x) y_s = f(x) $$

$ y_s $ is definitely a specific solution of $ (E) $.

Obvious solutions

In simple cases, we can directly try to find solutions, in specific by identifying the coefficients.

Examples of solving first order linear differential equation $ (LDE_1) $ with continuous coefficient

Here are several examples of how to resolve a $ EDL_1$ having the following general form:

$$ y' + a(x) y = f(x) \qquad (E) $$

With a constant coefficient

Let us take a constant coefficient $ a(x) = 2 $, and a function $ f(x) = x^2 $, then the equation $ (E) $ now becomes:

$$ y' + 2y = x^2 \qquad (E) $$

Solving the homogeneous equation $ (H) $

$$ y_h' + 2y_h = 0 \qquad (H) $$

$$ y_h'= -2y_h $$

$$ \frac{y_h'}{y_h} = -2 $$

$$ \int^x \frac{y_h'(t)}{y_h(t)} dt = \int^x -2dt $$

$$ ln|y_h| =-2x + K $$

$$ y_h= Ke^{- 2x } $$

We then have a homogeneous solution $ y_h $:

$$ y_h= Ke^{- 2x } $$

Solving the general equation $ (E) $

$$ y' + 2y = x^2 \qquad (E) $$

We are looking for a specific solution of the type:

$$ y_s= K(x)e^{- 2x } \qquad (y_s) $$

To do this, we can use the method of the variations of parameters, or use the coefficient identification method.

By the method of the variations of parameters

Using this method, we do have:

$$ y_s' + 2y_s = K'(x)e^{- 2x } - 2K(x)e^{- 2x } + 2 K(x)e^{- 2x } $$

$$ y_s' + 2y_s = K'(x)e^{- 2x } \qquad (E_p) $$

Given the following system:

$$ \Biggl \{ \begin{align*} y' + 2y = x^2 \qquad (E) \\ y_s' + 2y_s = K'(x)e^{- 2x } \qquad (E_p) \end{align*} $$

Thanks to both equations $ (E) $ and $ (E_p) $, we deduce an equation to solve, to determine $ K(x) $:

$$ K'(x)e^{- 2x } = x^2 $$

$$ K'(x) = x^2 e^{2x } $$

We antiderivate each side to determine $ K(x) $:

$$ K(x) = \int^x t^2 e^{2t } \hspace{0.1em} dt$$

We perform an integration by parts with the following choice of $ u $ and $ v' $:

$$ \Biggl \{ \begin{align*} u(t) = t^2 \\ v'(t) = e^{2t } \hspace{0.1em} dt \end{align*} $$

$$ \Biggl \{ \begin{align*} u'(t) = 2t \hspace{0.1em} dt \\ v(t) = \frac{1}{2} e^{2t } \end{align*} $$

$$K(x) = \Biggl[t^2 \frac{1}{2} e^{2t }\Biggr]^x - \int^x \frac{1}{2} 2t \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$

$$K(x) = x^2 \frac{1}{2} e^{2x }- \int^x t \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$

We perform another an integration by parts:

$$ \Biggl \{ \begin{align*} u(t) = t \\ v'(t) = e^{2t } \hspace{0.1em} dt \end{align*} $$

$$ \Biggl \{ \begin{align*} u'(t) = dt \\ v(t) = \frac{1}{2} e^{2t } \end{align*} $$

$$K(x) = x^2 \frac{1}{2} e^{2x }- \Biggl( \Biggl[t \frac{1}{2} e^{2t }\Biggr]^x - \int^x \frac{1}{2} \hspace{0.2em} e^{2t } \hspace{0.2em} dt \Biggr) $$

$$K(x) = x^2 \frac{1}{2} e^{2x }- x \frac{1}{2} e^{2x } + \int^x \frac{1}{2} \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$

$$K(x) = x^2 \frac{1}{2} e^{2x }- x \frac{1}{2} e^{2x } +\frac{1}{4} e^{2x } $$

$$K(x) = e^{2x } \Biggl( \frac{ x^2}{2}- \frac{x}{2} +\frac{1}{4}\Biggr) \qquad (K) $$

Now, we inject $ (K) $ into $ (y_s) $:

$$ y_s= e^{2x } \Biggl( \frac{ x^2}{2}- \frac{x}{2} +\frac{1}{4}\Biggr) e^{- 2x } $$

$$ y_s= \frac{ x^2}{2}- \frac{x}{2} +\frac{1}{4} $$

By the coefficient identification method

Given the relative simplicity of the equation $ (E) $:

$$ y' + 2y = x^2 \qquad (E) $$

We can look for a specific solution $ y_s $ of type second degree polynomial:

$$ y_s= ax^2 + bx +c \qquad (y_s) $$

By injecting $y_s$ into $ (E)$, we do have:

$$ (2ax + b) + 2(ax^2 + bx +c ) = x^2 $$

$$ 2ax^2 + (2a + 2b)x + (b+2c)= x^2 $$

Which leads us to the following system:

$$ \left \{ \begin{align*} 2a = 1 \\ 2a + 2b = 0 \\ b + 2c = 0 \end{align*} \right \} $$

$$ \left \{ \begin{align*} a = \frac{1}{2} \\ 1 + 2b = 0 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} b = -\frac{1}{2} \\ -\frac{1}{2} + 2c = 0 \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} c = \frac{1}{4} \end{align*} \right \} $$

So, we directly obtain solutions for $a, b, c$:

$$ \left \{ a = \frac{1}{2}, \hspace{0.2em} b = -\frac{1}{2}, \hspace{0.2em} c = \frac{1}{4} \right \} $$

And therefore a solution for $ y_s $:

$$ y_s= \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} $$

Addition of solutions

The total solution of $ (E) $ is the addition of both solutions $ y_h $ and $ y_s $, so:

$$ y_t= y_h + y_s $$

$$ y_t= Ke^{- 2x } + \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} $$

Verification of the total solution $ y_t$

If $ y_t $ is a solution of $ (E) $, then:

$$ y_t' + 2 y_t = x^2$$

Let us check it.

$$ y_t' + 2 y_t = \Biggl(-2 Ke^{- 2x } + x -\frac{1}{2} \Biggr) + 2 \Biggl( Ke^{- 2x } + \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} \Biggr)$$

$$ y_t' + 2 y_t = \hspace{0.2em} \underbrace{-2 Ke^{- 2x } + 2 Ke^{- 2x }} _\text{ $ = 0 $} \hspace{0.2em} + \hspace{0.2em} x^2 + \hspace{0.2em} \underbrace{ x -\frac{1}{2} -x + \frac{1}{2}} _\text{ $ = 0 $} $$

$$ y_t' + 2 y_t = x^2$$

$ y_t $ is definitely a total solution of $ (E) $.

With a continuous coefficient

Let us take a continuous coefficient $ a(x) = \frac{1}{x} $, and a function $ f(x) = x^3$, then the equation $ (E) $ becomes:

$$ y' + \frac{y}{x} = x^3 \qquad (E) $$

Solving the homogeneous equation $ (H) $

$$ y' + \frac{y}{x} = 0 \qquad (H) $$

$$ y' = - \frac{y}{x} $$

$$ \frac{y'}{y} = - \frac{1}{x} $$

$$ ln |y| = -ln|x| + C $$

$$ y = e^{-ln(x) + C} $$

$$ y = \frac{K}{x} $$

We then have a homogeneous solution $ y_h $:

$$ y_h= \frac{K}{x}$$

Solving the general equation $ (E) $

$$ y' + \frac{y}{x} = x^3 \qquad (E) $$

We are looking for a specific solution of the type:

$$ y_s= \frac{K(x)}{x} \qquad (y_s) $$

By the method of the variations of parameters

Using this method, we do have:

$$ y_s' + \frac{y}{x} = \frac {K'(x)x - K(x)}{x^2} + \frac {K(x) }{x^2} $$

$$ y_s' + \frac{y}{x} = \frac {K'(x)}{x} \qquad (E_p) $$

Given the following system:

$$ \left \{ \begin{align*} y' + \frac{y}{x} = x^3 \qquad \qquad (E) \\ y_s' + \frac{y_s}{x} = \frac {K'(x)}{x} \qquad (E_p) \end{align*} \right \} $$

Thanks to both equations $ (E) $ and $ (E_p) $, we deduce an equation to solve, to determine $ K(x) $:

$$\frac {K'(x)}{x} = x^3 $$

$$ K'(x) = x^4 $$

We antiderivate each side to determine $ K(x) $:

$$ K(x) = \int^x t^4 \hspace{0.2em} \hspace{0.1em} dt$$

$$ K(x) =\frac{x^5}{5} $$

Now, injecting $ (K) $ into $ (y_s) $:

$$ y_s= \frac{x^5}{5}\frac{1}{x} $$

$$ y_s= \frac{x^4}{5} $$

Addition of solutions

The total solution of $ (E) $ s the addition of the two solutions $ y_h $ and $ y_s $, so:

$$ y_t= y_h + y_s $$

$$ y_t= \frac{K}{x} + \frac{x^4}{5} $$

Verification of the total solution $ y_t$

If $ y_t $ is solution of $ (E) $, then:

$$ y_t' + \frac{y_t}{x} = x^3$$

Let us check it.

$$ y_t' + \frac{y_t}{x} = \Biggl(-\frac{ K }{x^2} + \frac{4x^3 }{5} \Biggr) + \Biggl( \frac{ \frac{K}{x} + \frac{x^4}{5}}{x}\Biggr)$$

$$ y_t' + \frac{y_t}{x} = -\frac{ K }{x^2} + \frac{4x^3 }{5} + \frac{K}{x^2} + \frac{x^4}{5x}$$

$$ y_t' + \frac{y_t}{x} = \hspace{0.2em} \underbrace{\frac{K}{x^2} - \frac{ K }{x^2}} _\text{ $= 0$} \hspace{0.2em} + \hspace{0.2em} \frac{4x^3 }{5} + \frac{x^3}{5}$$

$$ y_t' + \frac{y_t}{x} =x^3$$

$ y_t $ is definitely a total solution of $ (E) $.

Solving 1st order lin. diff. equations \( (LDE_1) \) with continuous coefficient

Demonstration

Solving the homogeneous equation \( (H) \)

Verification of the homogeneous solution \( y_h \)

Solving the general equation \( (E) \)

The method of the variations of parameters

Verification of the specific solution \( y_s \)

Obvious solutions

Examples of solving first order linear differential equation \( (LDE_1) \) with continuous coefficient

With a constant coefficient

Solving the homogeneous equation \( (H) \)

Solving the general equation \( (E) \)

By the method of the variations of parameters

By the coefficient identification method

Addition of solutions

Verification of the total solution \( y_t\)

With a continuous coefficient

Solving the homogeneous equation \( (H) \)

Solving the general equation \( (E) \)

By the method of the variations of parameters

Addition of solutions

Verification of the total solution \( y_t\)