Second Order Linear Differential Equation - Nonhomogeneous

Nonhomogeneous lineqr equations of second order with constant coefficients

Consider a nonhomgeneous equation of the form

$$L(y)=y''+ay'+by=R, \tag{1}$$

where the coeffieicnts $a$ and $b$ are constants, but $R$ is any function continuous on $(-\infty,\infty)$.

 

The general solution of the corresponding homogeneous equation $L(y)=0$ has the form $y_h=c_1v_1+c_2v_2,$ where

$$v_1(x)=e^{-ax/2}u_1(x), \hspace{1cm} v_2(x)=e^{-ax/2}u_2(x), \tag{2}$$

the functions $u_1$ and $u_2$ being determined by the discriminant of the equation: $d=a^2-4b.$

 

Theorem. Let $v_1$ and $v_2$ be the solutions of the equation $L(y)=0$ given by (2), where $L(y)=y''+ay'+by.$ Let $W$ denote the Wronskian of $v_1$ and $v_2.$ Then the non-homogeneous equation $L(y)=R$ has a particular solution $y_1$ given by the formula

$$y_1(x)=t_1(x)v_1(x)+t_2(x)v_2(x),$$

where

$$t_1(x)=-\int v_2(x)\frac{R(x)}{W(x)}dx, \hspace{1cm} t_2(x)=\int v_1(x) \frac{R(x)}{W(x)}. \tag{3}$$

 

Proof. Find functions $t_1$ and $t_2$ such that the combination $y_1=t_1v_1+t_2v_2$ will satisfy the equation $L(y)=R.$ We have

$$\begin{align*} y' &= t_1'v_1+t_1v_1'+t_2'v_2+t_2v_2' \\ &= t_1v_1' + t_2v_2' + (t_1'v_1+t_2'v_2) \\ y'' &=t_1'v_1'+t_1v_1''+t_2'v_2'+t_2v_2''+(t_1'v_1+t_2'v_2)' \\ &= t_1v_1''+t_2v_2''+(t_1'v_1'+t_2'v_2')+(t_1'v_1+t_2'v_2)'. \end{align*}$$

Then, we obtain the relation

$$\begin{align*} L(y_1) &= y_1''+ay_1'+by_1 \\ &= (v_1''+av_1'+bv_1)t_1 + (v_2'' + av_2' + bv_2)t_2 + (t_1'v_1'+t_2'v_2') + a(t_1'v_1+t_2'v_2) + (t_1'v_1+t_2'v_2)' \\ &= (t_1'v_1'+t_2'v_2')+(t_1'v_1+t_2'v_2)'+a(t_1'v_1+t_2'v_2) \\ & \hspace{3cm} \left( \text{Since } L(v_1)=L(v_2)=0 \right) \\ &= R. \end{align*} $$

This eqatuion is satisfied if

$$t_1'v_1+t_2'v_2=0 \hspace{1cm} \text{ and } \hspace{1cm} t_1'v_1'+t_2'v_2'=R.$$

In the matrix form, 

$$\begin{gather} \begin{pmatrix} v_1 & v_2 \\ v_1' & v_2' \end{pmatrix} \begin{pmatrix} t_1' \\ t_2' \end{pmatrix} = \begin{pmatrix} 0 \\ R \end{pmatrix}. \end{gather}$$

The determininant of the system is the Wronskian of $v_1$ and $v_2$. Since this is never zero, the system has a solution given by Cramer's rule:

$$t_1'=\frac{-v_2R}{W} \hspace{1cm} \text{ and } \hspace{1cm} t_2'=\frac{v_1R}{W}.$$

Integrating these relations, we obtain Eq. (3). This completes the proof.

 

Remark. Please note that the general solution of $L(y)=R$ is obtaine by adding to $y_1$ the general solution of the corresponding homogeneous equation, $y_h$: hence,

$$\begin{align*} y &= y_h+y_1 \\ &= c_1v_1+c_2v_2 + y_1 \end{align*} $$

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Find the general solution of the equation $y''-k^2y = R(x)$

 

The corresponding homogeneous equation has the general solution given by 

$$y_h=c_1e^{kx}+c_2e^{-kx}.$$

Then, particular solutions of the homogeneous equation are $v_1(x)=e^{kx}$ and $v_2(x)=e^{-kx}$, and their Wronskian is 

$$W(x)=v_1v_2'-v_1'v_2=e^{kx}(-ke^{-kx})-ke^{kx}(e^{-kx})=-2k.$$

We have the functions $t_1$ and $t_2$ given by

$$\begin{align*} t_1(x) &= -\int v_2 \frac{R}{W}dx \\ &= - \int e^{-kx} \frac{R(x)}{-2k}dx &= \frac{1}{2k} \int R(x)e^{-kx}dx \\  t_2(x) &= \int v_1 \frac{R}{W}dx \\ &= -\frac{1}{2k} \int R(x)e^{kx}dx. \end{align*}$$

Therefore, the particular solution $y_1$ of non-homogeneous equation is given by

$$ \begin{align*} y_1 &= t_1(x)v_1(x) + t_2(x)v_2(x) \\ &= \frac{e^{kx}}{2k} \int_0^x R(t)e^{-kt}dt - \frac{e^{-kx}}{2k} \int_0^x R(t)e^{kt}dt \\ &= \frac{1}{2k} \int_0^x R(t)e^{k(x-t)}dt - \frac{1}{2k} \int_0^x R(t)e^{-k(x-t)}dt \\ &= \frac{1}{k} \int_0^x R(t) \frac{e^{k(x-t)} - e^{-k(x-t)}}{2} dt \\ &= \frac{1}{k} \int_0^x R(t)\sinh{k(x-t)}dt. \end{align*} $$

Hence, for a nonzero constant $k$, the equation $y''-k^2y = R(x)$ has a particular solution $y_1$ given by

$$y_1 = \frac{1}{k} \int_0^x R(t)\sinh{k(x-t)}dt.$$

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Find the general solution of the equation $y''+k^2y=R(x)$

 

The corresponding homogeneous equation has the general solution given by

$$y_h(x)=c_1\cos{kx}+c_2\sin{kx}.\\

Particular solutions of $y_h$ are $v_1(x)=\cos{kx}$ and $v_2=\sin{kx},$ and the Wronskian of $v_1$ and $v_2$ is 

$$W(x)=v_1v_2'-v_1'v_2=\cos{kx} \left( k\cos{kx} \right) - \left( -k \sin{kx} \right) \sin{kx} = k.$$

The functions $t_1$ and $t_2$ are given by

$$\begin{align*} t_1(x) &= -\int \sin{kx} \frac{R(x)}{k} dx \\ &= -\frac{1}{k} \int R(x)\sin{kx}dx \\ t_2(x) &= \frac{1}{k} \int R(x)\cos{kx}. \end{align*} $$

Then, the particular solution $y_1$ of the non-homogeneous equation is given by

$$ \begin{align*} y_1 &= -\frac{\cos{kx}}{k} \int_0^x R(t) \sin{kt}dt + \frac{\sin{kx}}{k} \int_0^x R(t)\cos{kt}dt \\ &= \frac{1}{k} \int_0^x R(t) \left( \sin{kx} \cos{kt} - \cos{kx} \sin{kt} \right)dt \\ &= \frac{1}{k} \int_0^x R(t) \sin{k(x-t)}dt \\ & \hspace{3cm} \left( \text{Since } \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha} \sin{\beta} \right)\end{align*} $$

Therefore, if $k$ is a nonzero constant, the equation $y''+k^2y=R(x)$ has a particular solution $y_1$ given by

$$y_1=\frac{1}{k}\int_0^x R(t)\sin{k(x-t)}dt.$$