Linear differential equations of order 2 with constant coefficients 1

Form: \((E): a.y”(x)+b.y'(x) +cy(x) = f(x)\) where a,b,c are real constants \((a \neq 0)\) and \(f\) a function of the variable x

Equation without 2nd member

Let \((E_0)\) be the Diff. Eq. defined on the interval I by: \((E_0): a.y”(x)+b.y'(x)+c.y(x) = 0\) where a,b,c are real constants
By analogy with Diff. Eq. of the 1st order, we are looking for solutions of the form \(y(x)=e^{rx}\) where \(r\) is a constant to be determined.
2 real roots
If \((E_0)\) has 2 real roots \(r_1\) and \(r_2\), then the general solution of \((E_0)\) is :
\(y_0(x) = k_1.e^{r_1.x}+k_2 e^{r_2.x}\) where \(k_1\) and \(k_2\) are real constants.
Note: Special case where the characteristic equation: \(r^2-d=0\) with \(d \in \mathbb{R}^{*+}\), the general solution is: $$y_0(x)=k_1.e^{\sqrt{d}.x}+k_2.e^{-\sqrt{d}.x} = K_1.ch(\sqrt{d}.x)+K_2 .sh(\sqrt{d}.x)\text{ where }k_1,k_2,K_1,K_2 \in \mathbb{R}$$ 1 double root
If \((E_0)\) has 1 double root \(r_0\), then the general solution of \((E_0)\) is: \(y_0(x)=(k_1+k_2.x).e^{r_0.x}\) where \(k_1\) and \(k_2\) are real constants.
2 complex roots
If \((E_0)\) has 2 complex roots\(z_1=\alpha+i.\beta\) and \(z_2=\alpha-i.\beta\), then the general solution of \((E_0) \) is :
\(y_0(x)=[k_1.sin⁡(\beta.x)+k_2.cos⁡(\beta.x)] e^{\alpha.x}\) where \(k_1\) and \(k_2 \) are real constants.

Equation with 2nd member

We notice \(y_p(x)\) a particular solution of \((E)\).
\(\rightarrow \) All the solutions of \((E)\) on the interval I are the sum of the general solution and of the particular solution: \(y(x) = y_0(x)+y_p(x) \)

Finding a particular solution

We limit ourselves to the 2nd members \(f(x)\) of form \(f(x)=e^{rx}.P(x)\) where P(x) is a polynomial of degree \(n \in N\) and \(r\) a real
We consider:
$$(E): a.y^{”}(x)+b.y'(x) + c.y(x) = e^{r.x}.P(x)$$ We distinguish 3 cases:
If \(r\) not root of the characteristic equation then we seek 1 particular solution of form:
\(y_p(x)=e^{r.x}.Q(x)\) where \(Q(x)\) is a polynomial of degree n
If \(r\) a simple root of the characteristic equation then we seek 1 particular solution of form:
\(y_p(x)=x.e^{r.x}.Q(x)\) where \(Q(x)\) is a polynomial of degree n
If \(r\) a double root of the characteristic equation then we seek 1 particular solution of form:
\(y_p (x)=x^2 e^{r.x}.Q(x)\) where \(Q(x)\) is a polynomial of degree n
Example:
Let Diff. Eq. :
$$(E): 3.y^{”}(x) + y'(x) – 4.y(x) = x^2$$ Characteristic equation:
$$3.r^2 + r – 4 = 0$$ $$\Delta = b^2 – 4.a.c = 49$$ There are 2 roots: \(r_1=\frac{-4}{3}\) and \(r_2=1\)
From where:
$$y_0 (x)=k_1 e^{\frac{-4}{3}.x}+k_2.e^x$$ where \(k_1\) and \( k_2\) are real constants.
To determine the particular solution \(y_p(x)\), we are looking for 1 polynomial \(Q\) of degree 2 with the following general expression:
$$Q(x) = \alpha .x^2+\beta .x+\gamma$$ We obtain :
\(y’_p(x)=2.\alpha.x+\beta\) and \(y”_p(x)=2.\alpha\)
Substituting in \((E)\), we obtain the following system of equations:
$$\left\{ \begin{array}{ll} -4.\alpha = 1 \\ 2.\alpha-4.\beta = 0 \\ 6.\alpha + \beta – 4.\gamma = 0 \end{array} \right. \leftrightarrow \left\{ \begin{array}{ll} \alpha = \frac{-1}{4} \\ \beta = \frac{-1}{8} \\ \gamma = \frac{-13 }{32} \end{array} \right.$$ From where:
$$y_p(x)=-\frac{1}{4}.x^2-\frac{1}{8}.x-\frac{13}{32}$$ We deduce the general solution of the Diff. Eq. \((E)\):
$$y(x)=k_1.e^{\frac{-4}{3}.x}+k_2.e^x-\frac{1}{4}.x^2-\frac{1}{ 8}.x-\frac{13}{32}$$ where \(k_1\) and \(k_2\) are real constants.

Linear differential equations of order 2 with constant coefficients 2

Full form: \((E): a(x).y^{”}(x)+b(x).y'(x)+c(x).y(x)=d (x)\)
Associated homogeneous form: \((H): a(x).y^{”}(x)+b(x).y'(x)+c(x).y(x) = 0\)
For this to be a 2nd order equation, \(a(x) \neq 0 \rightarrow y^{”}=\frac{1}{a(x)}.(d(x)-b(x).y’ -c(x).y)\)
General solution = Homogeneous general solution + Particular solution

Principle of resolution

Resolving Diff. Eq. without 2nd member

We try to find a solution in exponential form: \(f(x)=e^{rx}\)
We have: \(f'(x)=r.e^{rx}\) and \(f^{”}(x)=r^2.e^{rx}\)
So: \(a.f^{”}(x)+b.f'(x)+c.f(x)=(a.r^2+b.r+c).e^{rx}\)
\(\rightarrow f(x)=e^{rx}\) is solution of H if and only if r root of (C): \(a.r^2+b.r+c=0\)
C is the Characteristic equation of the H
equation Let \(\delta=b^2-4.a.c\), be the discriminant of C:
If \(\delta > 0\), C admits 2 real roots: \(r_1=-b+\frac{\sqrt{\delta}}{2a}\) and \(r_2=-b-\frac {\sqrt{\delta}}{2a}\)
There are 2 exponential functions: \(f_1(x)= e^{r_1.x}\) and \(f_2(x)= e^{r_2.x}\), solutions of H
General solution of H: \(y=\lambda_1.e^{r_1.x}+\lambda_2.e^{r_2.x}\)
If \(\delta=0\), 1 solution: \(r=-b/2a\), so a solution of H: \(y(t)=(A.t+B)e^{-10t} \)
If \(\delta<0\), 2 complex conjugate solutions: \(r_1=-b+i\frac{\sqrt{\delta}}{2a} = \alpha+i\omega\) and \(r_2 =-b-i\frac{\sqrt{\delta}}{2a}=\alpha-i.\omega\)
With :\( \alpha=\frac{-b}{2a}\)and \(\omega=\frac{\sqrt{- \delta}}{2a}.\)

Solutions: \(f_1(x)= e^{r_1.x}\) and \(f_2(x)= e^{r_2.x} \rightarrow\) but problem because C unwieldy
So we have: \(g_1(x)=\frac{(f_1(x)+f_2(x))}{2}=e^{\alpha.x}.cos⁡(\omega.x)\) and \(g_2(x)=\frac{(f_1(x)-f_2(x))}{2.i}=e^{\alpha.x}.sin⁡(\omega.x)\) which are the real solutions of H

General Solution:
$$y=\lambda_1.g_1(x)+\lambda_2.g_2(x)=e^{\alpha.x}.[\lambda_1.cos⁡(\omega.x)+\lambda_2.sin⁡(\omega .x)]= e^{\alpha.x} [A.cos⁡(\omega.x)+B.sin⁡(\omega.x)]$$

Resolution with a particular solution (S.P.)

General guidelines for finding a particular solution (PS):
$$(E) a.y^{”}(x)+b.y'(x)+c.y(x)=d(x)$$ If \(d(x)\) 1 a polynomial of degree \(p\), then (E) admits 1 PS: \(a.Q^{”}+b.Q’+cQ=d(t)\) < br> Similarly degree \(p\), if \(Q\) is not root C \(\rightarrow\) if \(c \neq 0\)
Similarly degree \(p+1\), if \(Q\) 1 simple root of C \(\rightarrow\) if \(c = 0\) and \(b \neq 0\)
Similarly degree \(p+2\), if \(Q\) 1 double root of C \(\rightarrow\) if \(c=0\) and \(b=0\)
If \(d(x)\) of form \(P(x).e^{\lambda.x}\) (P(x) 1 a polynomial of degree \(p\)), then (E) admits 1 PS of form \(Q(x).e^{\lambda.x}\), \(Q(x)\) being 1 polynomial:
Similarly degree \(p\), if \(\lambda\) is not root C
Similarly degree \(p+1\), if \(\lambda\) simple root of C
Similarly degree \(p+2\), if \(\lambda\) double root of C
If \(d(x)\) of form \(\lambda.cos⁡(u.x)+\mu.sin⁡(u.x)\), then (E) admits 1 S.P.:
Of form \(A.cos⁡(u.x)+B.sin⁡(u.x)\)
Except if \(i.u\) root of C, then S.P.: \(x.(A.cos⁡(u.x)+B.sin⁡(u.x))\)

Method for varying constants \(\rightarrow\) when there is no indication
General solution of H: \(y=\lambda_1.f_1+\lambda_2.f_2\) with \(f_1\) and \(f_2\) 2 S.P.
We are looking for the general solution: \(y=f(x)=\lambda_1(x).f_1(x)+\lambda_2(x).f_2(x)\)
With \(y’=f'(x)=\lambda_1(x).f’_1(x)+\lambda_2(x).f’_2(x)\)
Linear system:
$$\left\{ \begin{array}{ll} f(x)=\lambda_1(x).f_1(x)+\lambda_2(x).f_2(x) \\ f'(x)=\lambda_1 (x).f’_1(x)+\lambda_2(x).f’_2(x) \end{array} \right. $$ $$\rightarrow \lambda’_1(x).f_1(x)+\lambda’_2(x).f_2(x)=0\text{ }(2)$$ We then have:
$$y^{”}=f^{”}(x)=\lambda’_1(x).f’_1(x)+\lambda_1(x).f^{”}_1(x)+\lambda’_2(x). f’_2(x)+\lambda_2(x).f^{”}_2(x)$$ We replace \(y\),\(y’\),\(y^{”}\) by the functions obtained, which gives:
$$a.\lambda’_1(x).f’_1(x)+a.\lambda’_2(x).f’_2(x) = d(x)$$ Combining with \((2)\),\( \lambda’_1\) and \(\lambda’_2\) system solutions:
$$\left\{ \begin{array}{ll} \lambda’_1(x).f_1(x)+\lambda’_2(x).f_2(x)=0 \\ \lambda’_1(x) .f’_1(x)+\lambda’_2(x).f’_2(x)=\frac{d(x)}{a} \end{array} \right. $$ We determine \(\lambda_1(x)\) and \(\lambda_2(x)\) by primitizing what we have just found.
Primitivation gives constants that correspond to the general solution of H.
Example:
Solve: \((E):y^{”}+y=\frac{1}{sin(x)}\) on \(I=]0;\pi[\)
\((H):y^{”}+y=0\) has the general solution: \(y=\lambda_1.cos(x)+\lambda_2.sin(x)\) (almost obvious because i root of the characteristic equation)
We are looking for the solutions of (E) in the form: \(y=\lambda_1.f_1(x)+\lambda_2.f_2(x)\) with \(f_1(x)=cos(x)\) and \(f_2(x)=sin(x)\).
We impose that: \(y’=f'(x)=\lambda_1(x).f’_1(x)+\lambda_2(x).f’_2(x)\), so that:
$$\lambda’_1(x).cos⁡(x)+\lambda’_2(x).sin(x)=0$$ We then have:
$$y^{”}=f^{”}(x)=\lambda’_1(x).f’_1(x)+\lambda_1(x).f^{”}_1(x)+\lambda’_2(x). f’_2(x)+\lambda_2(x).f^{”}_2(x)$$ So \(f\) solution of (E) when:
$$[\lambda’_1(x).f’_1(x)+\lambda_1(x).f^{”}_1(x)+\lambda’_2(x).f’_2(x)+\lambda_2( x).f^{”}_2(x)]+[\lambda_1.f_1(x)+\lambda_2.f_2(x)]=\frac{1}{sin(x)} $$ Using that \(f_1(x)\) and \(f_2(x)\) are solutions of (H), it do that:
$$\lambda’_1(x).f’_1(x)+\lambda’_2(x).f’_2(x)=\frac{1}{sin(x)} \leftrightarrow \lambda’_1( x).(-sin⁡(x))+\lambda’_2(x).cos⁡(x)=\frac{1}{sin(x)}$$ From where the system:
$$\left\{ \begin{array}{ll} \lambda_1.cos(x)+\lambda_2.sin(x)=0 \\ \lambda’_1(x).(-sin⁡(x))+ \lambda’_2(x).cos⁡(x)=\frac{1}{sin(x)} \end{array} \right. $$ We multiply the 1st equality by \(cos(x)\) and the 2nd by \((-sin(x))\) before adding to find: \(\lambda’_1(x)=-1\) so \(\lambda_1(x)=-x+C_1\)
We multiply the 1st equality by \(sin⁡(x)\) and the 2nd by \(cos(x)\) before adding to find: \(\lambda’_2(x)=\frac{cos(x)}{sin(x)}\) so \(\lambda_2(x)=ln(sin(x))+C_2\) (since \( sin(x)>0\)
The general solution of (E) is therefore: \(y=(-x+C_1).cos(x)+(ln(sin(x))+C_2).sin(x)\)
Or: \(y=(C_1.cos⁡(x)+C_2.sin⁡(x) )+(-x.cos⁡(x)+sin⁡(x).ln⁡(sin(x))) \ rightarrow\) to highlight the sum of a general solution of (H) and a PS of (E)
A particular solution of (E) is : \(f(x)=-x.cos⁡(x)+sin⁡(x).ln⁡(sin(x))\)
Example 2:
We consider the Diff. Eq. (Euler equation): \((E):x^2.y^{”}+x.y’+y=0\) \((x>0)\)
We introduce a new variable \(u\) by setting \(u=ln(x)\) for \(x>0\). We assume that \(y\) is a solution of (E).
Let \(z\) be the function of the variable \(u\), defined by: \(z(u)=z(ln(x))=y(x)=y(e^u)\)
1- Show the relations: \(y'(x)=z'(ln(x)).\frac{1}{x}\) and \(y^{”}(x)=z^{”}(ln(x )).\frac{1}{x^2}-z'(ln(x).\frac{1}{x^2}\)
Since \(y(x)=z.(ln(x))\), by differentiating, we obtain: \(y'(x)= z'(ln(x)).\frac{1}{x}\) and by differentiating a 2nd time, we obtain:
$$y^{”}(x)=-\frac{1}{x^2}.z'(ln(x))+z^{”}(ln(x)).\frac{1}{x}.\ frac{1}{x}=-z'(ln(x)).\frac{1}{x^2}+z^{”}(ln(x)).\frac{1}{x^2}$$ 2- Form a Diff. Eq. (E’) verified by \(z\)
Since \(y\) is assumed to be a solution of (E), we have:
$$x^2.y^{”}+x.y’+y=0 \leftrightarrow -z'(ln(x))+z^{”}(ln(x))+z'(ln(x))+z (ln(x))=0 \leftrightarrow z^{”}(u)+z(u)=0 \text{ }(E’)$$ 3) Solve \((E’)\) then \((E)\)
The solutions of \((E’)\) are the functions \(z\) of the type \(z(u)=A.cos⁡(u)+B.sin⁡(u)\), so coming back to the variable \(x\), there are the solutions of \((E)\) which are the functions:
$$y=y(x)=A.cos⁡(ln(x))+B.sin⁡(ln(x))$$