Yeah so far we consider the linear system of equations. Homogeneous system of equations. Say the x prime is equal to capital A times x where A is the n by n constant metrics right. Depending on the Eigen values of A and the corresponding Eigen vectors. The type of the general solution of this homogeneous system is determined right. We have studied how to define this general solution. Okay, now it's time to consider the corresponding non homogeneous problem. Okay, so from this one we added just a non homogeneous term plus the non homogeneous term which I denoted by capital F. Right then the general theory said that. Okay. Its general solution. Right. It's general solution Is of the form X is equal to excess of C process of P right. Where the X sub C is the so called the complementary solution. Okay, X sub C which is the complimentary solution which is a general solution of okay, this is the general solution of the corresponding homogeneous problem. Say X prime is equal to the A times X right, okay. And the X sub P this is any particular solution of okay Any particular solution of the original non homogeneous problem say X prime is equal to A times express of F. Right, okay. Before we have studied how to find the general solution of this, the homogeneous the problem. Right, okay. By analyzing the Eigen values of the coefficient matrix A and its corresponding Eigen back test right. Now it's time to concentrate on how to find the particular solution of this non homogeneous problem. Okay. We have a message a problem already in differential equations part 1 for the linear non homogeneous single equation. Okay, can you remind the following situation? In differential equation part 1 okay. We will consider the such the some problem of the type AY double prime, BY prime plus CY is equal to G of X right. Where G of X is not the necessarily to be 0. Okay, this is a non homogeneous problem. It's general solution Y equal to complementary solution plus particular solution Y sub P right. Okay. We have studied how to find the particular solution Y sub P right. Okay. There are two different cases. Right. Okay likewise for this non homogeneous system of equations we also have two different methods of finding a particular solution X sub P of this non homogeneous problem. Okay, that's that. This is something we will discuss from now on. Okay, so now let's consider this the non homogeneous problem. Okay, as in the case of the single linear non homogeneous differential equation, we also have two different methodists. The first one is a method of okay, method of undetermined coefficients. Okay, Undetermined the coefficients. Okay, As we already experienced from the case of the single differential equation, the method of undetermined coefficients okay, has some restriction. Okay, have some restriction. This method we can use the method of undetermined coefficients for some special type of the non homogeneous, term map. Right, okay. It is a pre cover Only when okay, this non homogeneous term is a polynomial. Okay, polynomial or the the exponential function. Okay. Right or the signs of co sines Right or sometimes the finite sums Or the products of these functions. Okay, That's the restriction. Okay but in case f is one of those the form. Okay. The polynomial or the exponential function or design or Cosines or finite sums of products of those functions. Okay, it is quite useful the math order to apply and each to compute the X sub P. Okay, I will explain that. Okay. Through the concrete examples right? Okay how to find the particular solution XP in those cases for example, let's think about the following problem. Okay, solve the initial value problem. Okay, X prime is equal to 1 minus 4 and 4 and 1 and times x plus okay, we have 3 plus T non homogeneous term minus 1 plus 4, t minus 10 E 33 T Okay. And the initial condition is X sub 0 is equal to 3 and 1. Okay, so this non homogeneous system of equations with initial conditions given by this one. Okay, again, before applying this initial condition right. We forced to find the general solution of that non homogeneous problem. Okay, it's general solution will be right. It's general solution As we know, X = Xc + Xp, okay? To find Xc, my complementary solution, we consider the corresponding homogeneous problem without this non-homogeneous term, okay? To solve this homogeneous problem, okay? We have to analyze the eigenvalues of eigenvectors of this coefficient matrix, right, okay? Let's call this coefficient matrix to be the matrix A, right? Then it's easy to find that, okay? Eigenvalues of A, okay? Are 1 +- 4i, okay? Its corresponding eigenvectors, okay? Corresponding eigenvectors are also given by the complex conjugated pairs, when I say 1, 0 -+ i times of vector 0, 1, okay? That's the corresponding eigenvectors, right? Then we can immediately write down the general solution of this corresponding the homogeneous problem, okay? Think about it, right, okay? I will skip several steps to reach to the X of c but you can check easily the following, right? The corresponding complementary solution Xc is given by, okay? So from using this information, okay? We get Xc = e to the t times, okay? Cosine 4t and sine 4t. I forgot the arbitrary constant of c sub 1, okay? In front let me put e sub t arbitrary = c1 times this column vector plus another arbitrary constant c2 times sine 4t, okay? And the minus cosine 4t, okay? That's a complementary solution, okay? So now it's remained to find this particular solution Xop, right, okay? The method of undetermined the coefficients suggest us to make a good guess for the form of the particular solution Xp by looking at this the non-homogeneous term, okay? Let's rely to this non-homogeneous term in the following equation, okay? What is non-homogeneous time F? You can write it as a? First you have a constant vector, right? 3-1+ you have t and 4t, that is a polynomial part, okay? So that 1, 4 and t, right? Then the exponential part that is a +0-10 and e to the 3t, right? You can decompose this non-homogeneous term into the term, okay? This is the constant part, this is the t part and this is exponential d part, right, okay? Looking at this the composition, right? Make a guess for this particular solution, okay? The basic rule to get the particular solution is, okay? For this type of the non-homogeneous problem, there should be a suitable particular solution which is of the same type as this to the f, okay? Except the coefficient part, right, okay? You must put the undetermined coefficients, okay? So what does that mean, right, okay? We make a guess for the Xp, right? We make the guess for the XP from this constant part. We make the constant and determine the coefficient vector a1, a2. From this polynomial part looking at this one, we make a guess b1, b2 times t, right? Undetermined coefficients, okay? And from the last to the exponential part, okay? You make a guess like c1, c2 and e to the 3t, okay? That's a guess, right? We make a guess for the particular solution Xp, right, okay? Plugging this into the differential equation, okay? And set the equality between the left hand side and the right hand side, okay? So let's go a few more steps, okay? Then what I need, okay? Let's make a copy of that equation here, okay? We have problem X prime =1-4 and 4, 1 and X +, for the F part you can write it as a 3-1 +1, 4 and t +0,-10 e to the 3t, okay? And I now make a guess for the type of the particular solution Xp in this version, okay? Lt's try, okay? Let's try that, our guess for X of p is really a particular solution of the original problem, okay? Then plugging this into the differential equation, what is the left hand side, right? Left hand side will be the derivative of Xp prime is equal to derivative of this is 0, derivative of this is just b1, b2, okay? Derivative of this will be 3[c1, c2] and e to the 3t, right, okay? That is equal to, right? That is equal to, we have 1, -4 and the 4, 1, okay? And the X of p that is equal to the a1, a2 and + b1, b2 and + c1, c2, I missed the t down there, okay? Let me see, there's a t + c1, c2, e to the 3t, okay? And +3, -1 and + 1, 4t + 0, -10 e to the 3t, okay? Okay, to the computation, right, for this part okay then you will get this is equal to first this is what? A1 minus 4a2 and the 4a1 plus a2, right, that's the question part, okay then the t part will be okay b1 minus 4b2. And the 4b1 plus b2 and the t plus exponential part you have 4c1 minus 4c2 and the 4c1 plus, no I made a mistake, what am I doing, right? This is not 4c1 but c1 minus 4c2, 4c1 plus c2 and the e to the 3t and plus 3 minus 1 and plus 1, 4 and plus 0 minus 10 and the e to 3t, right? Okay, so these two sides, right must be identically the same for all t, right? Okay, this is coming from this left hand side, right hand side of the equation, right, the differential system, right, right hand side, okay. Compare both sides, right this constant part must coincide with the constant part of the right hand side, what are they? We have a constant side question part is here and their way, so compare them. Okay so for example from this you will get, okay b1 must be equal to a1 minus 4 times 4a2, okay and plus 3, right? On the other hand this b2 must be equal to 4a1 plus a2, okay then minus 1, okay, that's the simultaneous equation. Okay and you also have coefficient of t, from the left hand side we don't have any time involving t, right? But on the right hand side we have a time involving, I'm sorry I forgot again here must be a t, right? Okay, down there so okay that means 0 is equal 2 for t part is b1 minus 4b2 plus 1, okay. And 0 is equal to 4b1 plus b2 plus 4, okay, that's the second equation. Okay, finally comparing the coefficient of e2 to 3t, okay, comparing the coefficient of e2 to 3t. You will have 3t1 is equal to was the coefficient of e to the 3t? Okay from here c1 minus 4c2 and plus 0, okay and the 3 times of c2 is equal to 4c1 plus c2 and minus 10. Okay, so we have a such the six simultaneous system of linear equations for six unknowns. A1, a2, b1, b2 and the c1, c2, right, okay, it's rather straightforward to solve this system of equations, okay. And then you will get okay, solving this simultaneous system of equations. My claim is the c1, c2 is equal to, okay, this is equal to 2 minus 1 and the b1 b2 must be equal to, okay. B1 b2 is equal to minus 1 and 0, okay and the a1 a2 must have been cool to 0 and 1, okay, that's my claim okay. It's easy matter, right to solve this six simultaneous system of equations for six unknowns, okay, what does that mean? Okay, so from this we have the excess of p, okay, that is equal to 0 1, okay and plus minus 1 0 t and plus 2 minus 1 e to the 3 t. Okay, that's the particular solution. Okay, so combining with the complementary solution we obtained already, right, we can claim that we have the general solution of the problem will be, right. Now, So General solution is x is equal to x of c plus x of p, okay we know that x of c, we obtained the x of c already, right? That is e to the t times c1 and cosine 4t, sine 4t and plus c2 sine 4t minus cosine 4t, right, that's a complimentary solution. Now, at this particular solution then, that is just read it 0 plus minus t, minus t plus 2 times of e to 3t. And the next entry will be 1 plus 0 minus e to 3t, right, that's the general solution of the original problem, okay. In fact, we start from the initial value problem, so using the initial conditions, okay, What was the initial condition? Let's look at the x0 is equal to the 1, okay, so from the initial condition, right from initial condition except 0, that is equal to 3 1, okay. The easy computation gives us that the c1 is equal we obtained okay, we get c1 is equal to 1 and the c2 is equal to minus 1. Okay, so that finally the solution of the initial value problem will be, right plug in c1 equal to 1 and the c2 is equal to negative 1 and add them up. Then x of t is equal to minus t plus e to the t, cosine 4tminus sine 4t plus 2 times the 3 to the 3t. And next entry will be 1 plus e to the t times sine 4t plus cosine 4t, Minus e to the 3t, right that's the solution to the given initial value problem, okay. [MUSIC]