This will greatly simplify the work required to find the coefficients. Given that \(y_p(x)=2\) is a particular solution to \(y3y4y=8,\) write the general solution and verify that the general solution satisfies the equation. Consider the nonhomogeneous linear differential equation, \[a_2(x)y+a_1(x)y+a_0(x)y=r(x). When this happens we just drop the guess thats already included in the other term. This will simplify your work later on. A particular solution for this differential equation is then. Embedded hyperlinks in a thesis or research paper, Counting and finding real solutions of an equation. We will justify this later. \nonumber \], \[\begin{align*}y+5y+6y &=3e^{2x} \\[4pt] (4Ae^{2x}+4Axe^{2x})+5(Ae^{2x}2Axe^{2x})+6Axe^{2x} &=3e^{2x} \\[4pt]4Ae^{2x}+4Axe^{2x}+5Ae^{2x}10Axe^{2x}+6Axe^{2x} &=3e^{2x} \\[4pt] Ae^{2x} &=3e^{2x}.\end{align*}\], So, \(A=3\) and \(y_p(x)=3xe^{2x}\). Notice that the second term in the complementary solution (listed above) is exactly our guess for the form of the particular solution and now recall that both portions of the complementary solution are solutions to the homogeneous differential equation. \[\begin{align*}x^2z_1+2xz_2 &=0 \\[4pt] z_13x^2z_2 &=2x \end{align*}\], \[\begin{align*} a_1(x) &=x^2 \\[4pt] a_2(x) &=1 \\[4pt] b_1(x) &=2x \\[4pt] b_2(x) &=3x^2 \\[4pt] r_1(x) &=0 \\[4pt] r_2(x) &=2x. If we multiplied the \(t\) and the exponential through, the last term will still be in the complementary solution. Also, we're using . There is nothing to do with this problem. Since the roots of the characteristic equation are distinct and real, therefore the complementary solution is y c = Ae -x + Be x Next, we will find the particular solution y p. For this, using the table, assume y p = Ax 2 + Bx + C. Now find the derivatives of y p. y p ' = 2Ax + B and y p '' = 2A . The characteristic equation for this differential equation and its roots are. The next guess for the particular solution is then. In other words we need to choose \(A\) so that. However, we wanted to justify the guess that we put down there. (Verify this!) Practice your math skills and learn step by step with our math solver. To use this online calculator for Complementary function, enter Amplitude of vibration (A), Circular damped frequency (d) & Phase Constant () and hit the calculate button. Boundary Value Problems & Fourier Series, 8.3 Periodic Functions & Orthogonal Functions, 9.6 Heat Equation with Non-Zero Temperature Boundaries, 1.14 Absolute Value Equations and Inequalities, \(A\cos \left( {\beta t} \right) + B\sin \left( {\beta t} \right)\), \(a\cos \left( {\beta t} \right) + b\sin \left( {\beta t} \right)\), \({A_n}{t^n} + {A_{n - 1}}{t^{n - 1}} + \cdots {A_1}t + {A_0}\), \(g\left( t \right) = 16{{\bf{e}}^{7t}}\sin \left( {10t} \right)\), \(g\left( t \right) = \left( {9{t^2} - 103t} \right)\cos t\), \(g\left( t \right) = - {{\bf{e}}^{ - 2t}}\left( {3 - 5t} \right)\cos \left( {9t} \right)\). PDF Second Order Differential Equations - University of Manchester Then, \(y_p(x)=u(x)y_1(x)+v(x)y_2(x)\) is a particular solution to the differential equation. \nonumber \], \[a_2(x)y+a_1(x)y+a_0(x)y=0 \nonumber \]. Differential Equations - Nonhomogeneous Differential Equations \end{align*}\], Note that \(y_1\) and \(y_2\) are solutions to the complementary equation, so the first two terms are zero. Which was the first Sci-Fi story to predict obnoxious "robo calls"? Connect and share knowledge within a single location that is structured and easy to search. However, we see that this guess solves the complementary equation, so we must multiply by \(t,\) which gives a new guess: \(x_p(t)=Ate^{t}\) (step 3). Notice that if we multiplied the exponential term through the parenthesis the last two terms would be the complementary solution. This is in the table of the basic functions. Writing down the guesses for products is usually not that difficult. The complementary solution is only the solution to the homogeneous differential equation and we are after a solution to the nonhomogeneous differential equation and the initial conditions must satisfy that solution instead of the complementary solution. So, differential equation will have complementary solution only if the form : dy/dx + (a)y = r (x) ? Learn more about Stack Overflow the company, and our products. So, with this additional condition, we have a system of two equations in two unknowns: \[\begin{align*} uy_1+vy_2 &= 0 \\[4pt] uy_1+vy_2 &=r(x). However, we will have problems with this. \[\begin{align*} a_1z_1+b_1z_2 &=r_1 \\[4pt] a_2z_1+b_2z_2 &=r_2 \end{align*}\], has a unique solution if and only if the determinant of the coefficients is not zero. Something seems wrong here. There are other types of \(g(t)\) that we can have, but as we will see they will all come back to two types that weve already done as well as the next one. Now, the method to find the homogeneous solution should give you the form The minus sign can also be ignored. Section 3.9 : Undetermined Coefficients. Our calculator allows you to check your solutions to calculus exercises. Now that weve gone over the three basic kinds of functions that we can use undetermined coefficients on lets summarize. Let's define a variable $u$ and assign it to the choosen part, Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. Then add on a new guess for the polynomial with different coefficients and multiply that by the appropriate sine. \(z_1=\frac{3x+3}{11x^2}\),\( z_2=\frac{2x+2}{11x}\), \[\begin{align*} ue^t+vte^t &=0 \\[4pt] ue^t+v(e^t+te^t) &= \dfrac{e^t}{t^2}. \(g\left( t \right) = 4\cos \left( {6t} \right) - 9\sin \left( {6t} \right)\), \(g\left( t \right) = - 2\sin t + \sin \left( {14t} \right) - 5\cos \left( {14t} \right)\), \(g\left( t \right) = {{\bf{e}}^{7t}} + 6\), \(g\left( t \right) = 6{t^2} - 7\sin \left( {3t} \right) + 9\), \(g\left( t \right) = 10{{\bf{e}}^t} - 5t{{\bf{e}}^{ - 8t}} + 2{{\bf{e}}^{ - 8t}}\), \(g\left( t \right) = {t^2}\cos t - 5t\sin t\), \(g\left( t \right) = 5{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\cos \left( {6t} \right) - \sin \left( {6t} \right)\), \(y'' + 3y' - 28y = 7t + {{\bf{e}}^{ - 7t}} - 1\), \(y'' - 100y = 9{t^2}{{\bf{e}}^{10t}} + \cos t - t\sin t\), \(4y'' + y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\), \(4y'' + 16y' + 17y = {{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right) + 6t\cos \left( {\frac{t}{2}} \right)\), \(y'' + 8y' + 16y = {{\bf{e}}^{ - 4t}} + \left( {{t^2} + 5} \right){{\bf{e}}^{ - 4t}}\). It is an exponential function, which does not change form after differentiation: an exponential function's derivative will remain an exponential function with the same exponent (although its coefficient might change due to the effect of the . I was just wondering if you could explain the first equation under the change of basis further. The condition for to be a particular integral of the Hamiltonian system (Eq. Did the Golden Gate Bridge 'flatten' under the weight of 300,000 people in 1987? Notice however that if we were to multiply the exponential in the second term through we would end up with two terms that are essentially the same and would need to be combined. Solve the complementary equation and write down the general solution. So this means that we only need to look at the term with the highest degree polynomial in front of it. This is not technically part the method of Undetermined Coefficients however, as well eventually see, having this in hand before we make our guess for the particular solution can save us a lot of work and/or headache. First, we will ignore the exponential and write down a guess for. Complementary function Calculator | Calculate Complementary function 0.00481366327239356 Meter -->4.81366327239356 Millimeter, Static Force using Maximum Displacement or Amplitude of Forced Vibration, Maximum Displacement of Forced Vibration using Natural Frequency, Maximum Displacement of Forced Vibration at Resonance, Maximum Displacement of Forced Vibration with Negligible Damping, Total displacement of forced vibration given particular integral and complementary function, The Complementary function formula is defined as a part of the solution for the differential equation of the under-damped forced vibrations and is represented as, The Complementary function formula is defined as a part of the solution for the differential equation of the under-damped forced vibrations is calculated using. What is the solution for this particular integral (ODE)? Interpreting non-statistically significant results: Do we have "no evidence" or "insufficient evidence" to reject the null? So when \(r(x)\) has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. But when we substitute this expression into the differential equation to find a value for \(A\),we run into a problem. This means that if we went through and used this as our guess the system of equations that we would need to solve for the unknown constants would have products of the unknowns in them. This one can be a little tricky if you arent paying attention. y +p(t)y +q(t)y = g(t) (1) (1) y + p ( t) y + q ( t) y = g ( t) where g(t) g ( t) is a non-zero function. \end{align*}\], \[y(t)=c_1e^{3t}+c_2+2t^2+\dfrac{4}{3}t.\nonumber \]. Or. e^{2x}D(e^{-2x}(D - 3)y) & = e^{2x} \\ \end{align*} \nonumber \], So, \(4A=2\) and \(A=1/2\). Plugging this into our differential equation gives. Lets take a look at the third and final type of basic \(g(t)\) that we can have. This will arise because we have two different arguments in them. $$ Integral Calculator With Steps! \nonumber \] So, we have an exponential in the function. Now, apply the initial conditions to these. and as with the first part in this example we would end up with two terms that are essentially the same (the \(C\) and the \(G\)) and so would need to be combined. This is a case where the guess for one term is completely contained in the guess for a different term. The more complicated functions arise by taking products and sums of the basic kinds of functions. All common integration techniques and even special functions are supported. Again, lets note that we should probably find the complementary solution before we proceed onto the guess for a particular solution. One of the nicer aspects of this method is that when we guess wrong our work will often suggest a fix. Second Order Differential Equation - Solver, Types, Examples - Cuemath The complementary equation is \(yy2y=0\), with the general solution \(c_1e^{x}+c_2e^{2x}\). But that isnt too bad. So, we would get a cosine from each guess and a sine from each guess. This is because there are other possibilities out there for the particular solution weve just managed to find one of them. Notice two things. My answer assumes that you know the full proof of the general solution of homogenous linear ODE. (D - 2)^2(D - 3)y = 0. D(e^{-3x}y) & = xe^{-x} + ce^{-x} \\ So, in general, if you were to multiply out a guess and if any term in the result shows up in the complementary solution, then the whole term will get a \(t\) not just the problem portion of the term. Notice that even though \(g(t)\) doesnt have a \({t^2}\) in it our guess will still need one! \nonumber \], Find the general solution to \(y4y+4y=7 \sin t \cos t.\). Complementary function / particular integral. 15 Frequency of Under Damped Forced Vibrations Calculators. In these solutions well leave the details of checking the complementary solution to you. My text book then says to let y = x e 2 x without justification. With only two equations we wont be able to solve for all the constants. Generic Doubly-Linked-Lists C implementation. Once the problem is identified we can add a \(t\) to the problem term(s) and compare our new guess to the complementary solution. If so, multiply the guess by \(x.\) Repeat this step until there are no terms in \(y_p(x)\) that solve the complementary equation. We need to calculate $du$, we can do that by deriving the equation above, Substituting $u$ and $dx$ in the integral and simplify, Take the constant $\frac{1}{5}$ out of the integral, Apply the integral of the sine function: $\int\sin(x)dx=-\cos(x)$, Replace $u$ with the value that we assigned to it in the beginning: $5x$, Solve the integral $\int\sin\left(5x\right)dx$ and replace the result in the differential equation, As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$. Then tack the exponential back on without any leading coefficient. \end{align*}\], Applying Cramers rule (Equation \ref{cramer}), we have, \[u=\dfrac{\begin{array}{|lc|}0 te^t \\ \frac{e^t}{t^2} e^t+te^t \end{array}}{ \begin{array}{|lc|}e^t te^t \\ e^t e^t+te^t \end{array}} =\dfrac{0te^t(\frac{e^t}{t^2})}{e^t(e^t+te^t)e^tte^t}=\dfrac{\frac{e^{2t}}{t}}{e^{2t}}=\dfrac{1}{t} \nonumber \], \[v= \dfrac{\begin{array}{|ll|}e^t 0 \\ e^t \frac{e^t}{t^2} \end{array} }{\begin{array}{|lc|}e^t te^t \\ e^t e^t+te^t \end{array} } =\dfrac{e^t(\frac{e^t}{t^2})}{e^{2t}}=\dfrac{1}{t^2}\quad(\text{step 2}). To do this well need the following fact. (You will get $C = -1$.). We now want to find values for \(A\), \(B\), and \(C\), so we substitute \(y_p\) into the differential equation. It is now time to see why having the complementary solution in hand first is useful. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by \(x\). \end{align*}\], Then,\[\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}=\begin{array}{|ll|}x^2 2x \\ 1 3x^2 \end{array}=3x^42x \nonumber \], \[\begin{array}{|ll|}r_1 b_1 \\ r_2 b_2 \end{array}=\begin{array}{|ll|}0 2x \\ 2x -3x^2 \end{array}=04x^2=4x^2. The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of \(r(x)\). The complementary equation is \(y2y+y=0\) with associated general solution \(c_1e^t+c_2te^t\). Based on the form of \(r(x)\), we guess a particular solution of the form \(y_p(x)=Ae^{2x}\). Now, tack an exponential back on and were done. Conic Sections Transformation. \[y_p(x)=3A \sin 3x+3B \cos 3x \text{ and } y_p(x)=9A \cos 3x9B \sin 3x, \nonumber \], \[\begin{align*}y9y &=6 \cos 3x \\[4pt] 9A \cos 3x9B \sin 3x9(A \cos 3x+B \sin 3x) &=6 \cos 3x \\[4pt] 18A \cos 3x18B \sin 3x &=6 \cos 3x. In fact, if both a sine and a cosine had shown up we will see that the same guess will also work. Particular Integral - Where am i going wrong!? Remembering to put the -1 with the 7\(t\) gives a first guess for the particular solution. The complementary solution this time is, As with the last part, a first guess for the particular solution is. The way that we fix this is to add a \(t\) to our guess as follows. What does to integrate mean? Find the general solutions to the following differential equations. We can only combine guesses if they are identical up to the constant. At this point the reason for doing this first will not be apparent, however we want you in the habit of finding it before we start the work to find a particular solution. \end{align*}\]. I hope they would help you understand the matter better. However, even if \(r(x)\) included a sine term only or a cosine term only, both terms must be present in the guess. Okay, we found a value for the coefficient. y & = -xe^{2x} + Ae^{2x} + Be^{3x}. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here. C.F. Improving the copy in the close modal and post notices - 2023 edition, New blog post from our CEO Prashanth: Community is the future of AI, Particular Integrals for Second Order Differential Equations with constant coefficients. Substitute \(y_p(x)\) into the differential equation and equate like terms to find values for the unknown coefficients in \(y_p(x)\). What was the actual cockpit layout and crew of the Mi-24A? We will ignore the exponential and write down a guess for \(16\sin \left( {10t} \right)\) then put the exponential back in. This page titled 17.2: Nonhomogeneous Linear Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin Jed Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. 18MAT21 MODULE. If total energies differ across different software, how do I decide which software to use? It helps you practice by showing you the full working (step by step integration). We now need move on to some more complicated functions. When this happens we look at the term that contains the largest degree polynomial, write down the guess for that and dont bother writing down the guess for the other term as that guess will be completely contained in the first guess. Are there any canonical examples of the Prime Directive being broken that aren't shown on screen? (6.26)) is symmetrical with respect to and H. Therefore, if a bundle defined by is a particular integral of a Hamiltonian system with function H, then H is also a particular integral of a Hamiltonian system with function . 2.9: Integrals Involving Exponential and Logarithmic Functions In this section we will take a look at the first method that can be used to find a particular solution to a nonhomogeneous differential equation. If we get multiple values of the same constant or are unable to find the value of a constant then we have guessed wrong. Upon doing this we can see that weve really got a single cosine with a coefficient and a single sine with a coefficient and so we may as well just use. As we will see, when we plug our guess into the differential equation we will only get two equations out of this. The guess for the \(t\) would be, while the guess for the exponential would be, Now, since weve got a product of two functions it seems like taking a product of the guesses for the individual pieces might work. Solving this system gives \(c_{1} = 2\) and \(c_{2} = 1\). This reasoning would lead us to the . Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor . From our previous work we know that the guess for the particular solution should be. The two terms in \(g(t)\) are identical with the exception of a polynomial in front of them. Circular damped frequency refers to the angular displacement per unit time. Accessibility StatementFor more information contact us atinfo@libretexts.org. The nonhomogeneous equation has g(t) = e2t. Solve the following differential equations a) (D-3D2+3D - Dx=e* +2. The difficulty arises when you need to actually find the constants. \nonumber \], When \(r(x)\) is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. This would give. So, what went wrong? Solving a Second-Order Linear Equation (Non-zero RHS), Questions about auxiliary equation and particular integral. Particular integral and complementary function - Math Theorems To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation. Check out all of our online calculators here! \nonumber \], \[z2=\dfrac{\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}=\dfrac{2x^3}{3x^42x}=\dfrac{2x^2}{3x^3+2}.\nonumber \], \[\begin{align*} 2xz_13z_2 &=0 \\[4pt] x^2z_1+4xz_2 &=x+1 \end{align*}\]. Complementary function Definition & Meaning - Merriam-Webster Now, back to the work at hand. However, upon doing that we see that the function is really a sum of a quadratic polynomial and a sine. Taking the complementary solution and the particular solution that we found in the previous example we get the following for a general solution and its derivative. One final note before we move onto the next part. In this case the problem was the cosine that cropped up. This first one weve actually already told you how to do. The guess here is. EDIT A good exercice is to solve the following equation : Here the emphasis is on using the accompanying applet and tutorial worksheet to interpret (and even anticipate) the types of solutions obtained. Ify1(x)andy2(x)are any two (linearly independent) solutions of a linear, homogeneous second orderdierential equation then the general solutionycf(x),is ycf(x) =Ay1(x) +By2(x) whereA, Bare constants. Consider the following differential equation | Chegg.com So, we need the general solution to the nonhomogeneous differential equation. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Parametric Equations and Polar Coordinates, 9.5 Surface Area with Parametric Equations, 9.11 Arc Length and Surface Area Revisited, 10.7 Comparison Test/Limit Comparison Test, 12.8 Tangent, Normal and Binormal Vectors, 13.3 Interpretations of Partial Derivatives, 14.1 Tangent Planes and Linear Approximations, 14.2 Gradient Vector, Tangent Planes and Normal Lines, 15.3 Double Integrals over General Regions, 15.4 Double Integrals in Polar Coordinates, 15.6 Triple Integrals in Cylindrical Coordinates, 15.7 Triple Integrals in Spherical Coordinates, 16.5 Fundamental Theorem for Line Integrals, 3.8 Nonhomogeneous Differential Equations, 4.5 Solving IVP's with Laplace Transforms, 7.2 Linear Homogeneous Differential Equations, 8. As mentioned prior to the start of this example we need to make a guess as to the form of a particular solution to this differential equation. Notice that in this case it was very easy to solve for the constants. This is especially true given the ease of finding a particular solution for \(g\)(\(t\))s that are just exponential functions. For \(y_p\) to be a solution to the differential equation, we must find a value for \(A\) such that, \[\begin{align*} yy2y &=2e^{3x} \\[4pt] 9Ae^{3x}3Ae^{3x}2Ae^{3x} &=2e^{3x} \\[4pt] 4Ae^{3x} &=2e^{3x}. We have \(y_p(t)=2At+B\) and \(y_p(t)=2A\), so we want to find values of AA and BB such that, Solve the complementary equation and write down the general solution \[c_1y_1(x)+c_2y_2(x). \(y(t)=c_1e^{2t}+c_2te^{2t}+ \sin t+ \cos t \). The second and third terms are okay as they are. In other words, we had better have gotten zero by plugging our guess into the differential equation, it is a solution to the homogeneous differential equation! Integration is a way to sum up parts to find the whole. Differential Equations Calculator & Solver - SnapXam The complementary equation is \(y+4y+3y=0\), with general solution \(c_1e^{x}+c_2e^{3x}\). When solving ordinary differential equation, why use specific formula for particular integral. Complementary function Calculator | Calculate Complementary function We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation. Sometimes, \(r(x)\) is not a combination of polynomials, exponentials, or sines and cosines. It only takes a minute to sign up. Substitute back into the original equation and solve for $C$. How to combine several legends in one frame? Here is how the Complementary function calculation can be explained with given input values -> 4.813663 = 0.01*cos(6-0.785398163397301). Is it safe to publish research papers in cooperation with Russian academics? We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters. Solving this system of equations is sometimes challenging, so lets take this opportunity to review Cramers rule, which allows us to solve the system of equations using determinants. Dipto Mandal has verified this Calculator and 400+ more calculators! This still causes problems however. The first two terms however arent a problem and dont appear in the complementary solution. Now, lets proceed with finding a particular solution. It's not them. $$ The guess for the polynomial is. rev2023.4.21.43403. We have, \[y(x)=c_1 \sin x+c_2 \cos x+1 \nonumber \], \[y(x)=c_1 \cos xc_2 \sin x. Complementary function / particular integral - Mathematics Stack Exchange For this example, \(g(t)\) is a cubic polynomial. There a couple of general rules that you need to remember for products. Lets take a look at another example that will give the second type of \(g(t)\) for which undetermined coefficients will work. We have one last topic in this section that needs to be dealt with. Then, \(y_p(x)=(\frac{1}{2})e^{3x}\), and the general solution is, \[y(x)=c_1e^{x}+c_2e^{2x}+\dfrac{1}{2}e^{3x}. Upon multiplying this out none of the terms are in the complementary solution and so it will be okay. Another nice thing about this method is that the complementary solution will not be explicitly required, although as we will see knowledge of the complementary solution will be needed in some cases and so well generally find that as well. Solve a nonhomogeneous differential equation by the method of undetermined coefficients. Solving this system gives us \(u\) and \(v\), which we can integrate to find \(u\) and \(v\). We know that the general solution will be of the form. In this case, the solution is given by, \[z_1=\dfrac{\begin{array}{|ll|}r_1 b_1 \\ r_2 b_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}} \; \; \; \; \; \text{and} \; \; \; \; \; z_2= \dfrac{\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}. \nonumber \]. The guess for this is then, If we dont do this and treat the function as the sum of three terms we would get. Step 3: Finally, the complementary angle for the given angle will be displayed in the output field. Thus, we have, \[(uy_1+vy_2)+p(uy_1+vy_2)+(uy_1+vy_2)=r(x). Solve the following initial value problem using complementary function and particular integral method( D2 + 1)y = e2* + cosh x + x, where y(0) = 1 and y'(o) = 2 a) Q2. So, differentiate and plug into the differential equation. Let \(y_p(x)\) be any particular solution to the nonhomogeneous linear differential equation, Also, let \(c_1y_1(x)+c_2y_2(x)\) denote the general solution to the complementary equation.