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For the linear differential operator $L=(D-\alpha )(D-\beta )$, the exponential function is an eigenfunction: $L{e}^{ϵx}=(ϵ-\alpha )(ϵ-\beta )e^{ϵx}$. Distinct $\alpha ,\beta$ give the two homogeneous solutions, while the method of undetermined coefficients supplies a particular solution at the distinct forcing exponent $\gamma$:

For $x>0$, use the Euler differential equation substitution $t=\ln x$, so $x=e^t$, $x{y}^{\prime }=\dot{y}$ and $x^2{y}^{\prime \prime }=\ddot{y}-\dot{y}$. The transformed equation isIts characteristic polynomial factors as $(r-1)(r+2)$. The forcing $e^t$ coincides with a homogeneous solution, so the method of undetermined coefficients requires a particular solution $y_p=At{e}^t$. Substitution gives $3A{e}^t=e^t$, hence $A=1/3$. Transforming back,The conditions at $x=1$ give $C_1+C_2=0$ and $C_1-2C_2+1/3=0$, soDifferentiating confirms both initial values and the original equation. The solution interval containing $x=1$ excludes the singular point $x=0$.

The characteristic equation $r^2+5r+4=(r+1)(r+4)=0$ gives the complementary function $C_1{e}^{-x}+C_2{e}^{-4x}$. Write the forcing as $30e^x+30e^{-x}$. The method of undetermined coefficients gives $3e^x$ for the first term. Because $e^{-x}$ already belongs to the homogeneous solution, resonance in a differential equation requires a trial $Kx{e}^{-x}$ for the second; substitution gives $3K=30$. Therefore

Use the method of undetermined coefficients with the polynomial $f_p=a{t}^3+b{t}^2+ct+d$. Substitution into $f_p^{\prime \prime }+3f_p^{\prime }+2f_p$ givesMatching successive coefficients with the forcing gives $a=1$, $b=-3$, $c=4$ and $d=1$. Hence a particular integral isIndeed $f_p^{\prime }=3t^2-6t+4$ and $f_p^{\prime \prime }=6t-6$ give $f_p^{\prime \prime }+3f_p^{\prime }+2f_p=2t^3+3t^2-4t+8$. The complementary function, if wanted, is $C_1{e}^{-t}+C_2{e}^{-2t}$.

The complementary solution comes from the characteristic equation $r^4-a^4=0$, whose roots are $a,-a,ia,-ia$. It is thereforeFor the resonant exponential particular solution, the forcing is already a homogeneous exponential, so the method of undetermined coefficients needs the resonant trial $y_p=Kx{e}^{-ax}$. If $P(r)=r^4-a^4$, the identity $P(D)[x{e}^{rx}]=P^{\prime }(r)e^{rx}$ when $P(r)=0$ gives $P^{\prime }(-a)=-4a^3$. Hence $K=-1/(4a^3)$ andThe decay condition first eliminates the growing exponential, $C_1=0$. A nonzero combination of the sine and cosine has a fixed nonzero amplitude and cannot tend to zero, so $C_3=C_4=0$ as well. Both remaining exponential terms decay because $a>0$. Finally $y(0)=C_2=1$, giving the unique answerThis is an instance of resonance in a differential equation: multiplication by $x$ supplies the particular solution when the forcing exponent is a simple characteristic root.

The characteristic roots of the homogeneous linear ordinary differential equation are $2$ and $3$. The forcing is resonant with the $e^{3x}$ solution, so the method of undetermined coefficients uses the particular solution $x{e}^{3x}$. Thus $y=A{e}^{2x}+B{e}^{3x}+x{e}^{3x}$. The two initial conditions are $A+B=0$ and $2A+3B+1=0$, giving

The characteristic polynomial factors as $(r-2)(r+1)$, so the complementary solution is $A{e}^{2t}+B{e}^{-t}$. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses $t{e}^{2t}$ and $t{e}^{-t}$. Substitution gives the particular solution $t{e}^{2t}-t{e}^{-t}-3t$. ThusThe initial conditions give $A+B=0$ and $2A-B-3=0$. HenceThe factors of $t$ in the particular solution are essential: an unmodified exponential trial lies in the kernel of the differential operator.

Use the method of undetermined coefficients. Substitution of $y_p=(a+bx)\sin x+(c+dx)\cos x$ givesEquating the coefficients of $x\sin x$, $x\cos x$, $\sin x$, and $\cos x$ to the forcing gives $d=1$, $b=0$, $c=1$, $a=-1$. ThusThe characteristic equation of the homogeneous equation is $(r-1{)}^2=0$, with a repeated root. Its complementary function is $(C_1+C_{2}x)e^x$. Since $y_p(0)=1$ and $y_p^{\prime }(0)=0$, the initial conditions give $C_1=-1$ and $C_2=1$. ThereforeDirect differentiation verifies both zero initial data and the original forcing.

The characteristic polynomial factors as $(r+1)(r+2)$, so the homogeneous solution is $A{e}^{-x}+B{e}^{-2x}$. The forcing $e^{-x}$ already occurs in that solution, so a constant multiple of $e^{-x}$ cannot be a particular solution. Use $y_p=Cx{e}^{-x}$ instead. Its derivatives are $C(1-x)e^{-x}$ and $C(x-2)e^{-x}$, so substitution gives $C{e}^{-x}$ and fixes $C=1$. ThusThe extra factor of $x$ accounts for the repeated forcing exponent in the method of undetermined coefficients.

For the homogeneous linear differential equation, the characteristic equation is $r^2-2r+2=0$, with polynomial roots $1\pm i$. Its homogeneous solution is therefore $e^x(A\cos x+B\sin x)$.

Use the method of undetermined coefficients for a particular solution $y_p=a{x}^2+bx+c$. Substitution givesEquating coefficients yields $a=1$, $b=2$ and $c=1$. Hence the general solution is

Use the method of undetermined coefficients with a particular solution $K{e}^{-\sqrt{3}x}$. The differential operator multiplies this exponential function by $3-3+3=3$. Matching the forcing coefficient gives $3K=1$, hence

Use the method of undetermined coefficients, choosing a particular solution $y_p=A+B\cos 2x$. Substitution givesMatching the constant and cosine coefficients to $\frac{1}{2}-\frac{1}{2}\cos 2x$ gives $A=1/2$, $B=1/6$. ThusNeither forcing term is resonant with the homogeneous angular frequency one.

The homogeneous solution is $A\cos x+B\sin x$. If $k\ne 1$, the method of undetermined coefficients permits a particular integral $C\cos kx$. Substitution gives $(1-k^2)C=1$ and henceFor $k=1$, the forcing is itself a homogeneous solution, so this trial fails by resonance in a differential equation. Multiplying the sine trial by $x$ givesThus the resonant general solution isThe secular term containing $x$ describes the increasing amplitude under undamped resonant forcing.

The associated homogeneous linear differential equation has characteristic roots $2\pm i$, so its complementary function is $e^{2x}(A\cos x+B\sin x)$. For a particular integral, the method of undetermined coefficients suggests $y_p=a{x}^2+bx+c$. Substitution and comparison of coefficients givesso $y_p=25x^2+40x+22$. The general solution is thereforeThe first boundary condition gives $A=-21$. At $x=\pi /2$, the polynomial part already equals the assigned boundary value and the cosine vanishes, so $e^{\pi }B=0$. ThusThis is the unique solution satisfying the two boundary conditions; here the phrase “particular solution” means that selected member of the full family, rather than only the polynomial particular integral.

The characteristic equation of a constant-coefficient differential equation here is $r^2-3r+2=(r-1)(r-2)=0$. Its homogeneous solution is $C_1{e}^x+C_2{e}^{2x}$. Because the forcing $e^x$ is already a homogeneous mode, the method of undetermined coefficients requires $y_p=Ax{e}^x$ rather than $A{e}^x$.

Here $y_p^{\prime }=A(x+1)e^x$ and $y_p^{\prime \prime }=A(x+2)e^x$, so $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=-A{e}^x$. Thus $A=-1$, and the general solution is

For the forcing $8e^{-2x}$, the method of undetermined coefficients gives the particular term $2e^{-2x}$. The term $e^{-x}$ is resonant with a homogeneous solution, so use $ax{e}^{-x}$ for its forcing; substitution gives $a=1$.

Thus $y=C_1{e}^{2x}+C_2{e}^{-x}+2e^{-2x}+x{e}^{-x}$. The two initial conditions reduce to $C_1+C_2=3$ and $2C_1-C_2=0$, so $C_1=1$, $C_2=2$. The solution of the initial value problem isThe extra factor $x$ in the resonant particular solution is essential.