30 Differential equation of first order 1 Part 2

Added: 2026-04-29

[00:00:00]so after formation of differential equation now we are going to talk about solution of differential equation so solving a differential equation simply means finding a function which will satisfy the given differential equation so here we have one problem the question is verify that this function e to the power 2 X this is nothing but Y y equals 2 e to the power 2 X is the solution of this differential equation so as I have said earlier the soleus solving a differential equation means finding this function y which will satisfy this differential equation so here we only have to verify that this is the solution of given differential equation so y is e to the power 2 X now what will be the value of dy by DX dy by DX will be nothing but 2 into e to the power 2 X and D 2 y by DX square we again have to differentiate this so we will hit 4 into e to the power 2 X now we have to show that for this function the left hand side will be equal to right hand side see there is one a very basic concept that you need to understand suppose if I have this equation X plus 2 equals to 5 then what is the solution of this equation the solution is x equals to 3 y x equals to 3 in solution because when X is 3 your left hand side will be 3 plus 2 5 this is equal to right hand side so if I substitute the value of variable here then the left hand side becomes equal to right hand side and therefore this well particular value of the variable is the solution of this equation so similarly here if I have to verify that this y equals 2 e to the power 2 X is solution of this differential equation then for this given function we have to so that left hand side which is this expression will be equal to the right hand side to 0 so if Y is e to the power 2 X then our left hand side will be left hand side is d 2 y by DX s square minus 3 dy by DX plus a2y now in place of why we have to substitute e to the power 2 X dy by DX will be this thing D 2 y by DX square will be 4 into a to the power 2 X so this will be 4 e to the power 2 X minus 3 dy by DX s2 + 2 e to the power 2 X plus 2y is e to the power 2 X so now if you see this the left hand side is nothing but 4 e to the power 2 X minus 3 and todo 6 e to the power 2 X plus 2 into e to the power 2 X so finally this will give us 4 + 2 6 - 6 0 which was our right hand side so since for the given function the left hand side becomes equal to right hand side therefore this function is the solution of this differential equation so this question was basically how to find whether a function is solution of given differential equation or not ok now we are going to learn how to find this solution how to solve a differential equation how to get this function Y ok so we'll start with the solution of differential equation of first order and first degree there will see 4 different techniques and then we'll talk about differential equation of higher order okay so next we have to learn solution of differential equation of first order in first degree in our basic disks in our idea this person we have seen what is the meaning of first order and first degree if only we have dy by DX then it means the differential equation will be for first order and if the exponent of dy by DX is 1 then it will be a first degree differential equation now for these type of differential equation there are different techniques to solve this type of differential equation so that we are going to see one by one our first method will be variable separable method ok so this will be our first method then homogeneous differential equation linear differential equation obviously the India differential equation of first order and first degree we are only talking about differential equation of first order first degree so homogeneous then linear differential equation we will talk about Bernoulli differential equation now this Bernoulli differential equation is not linear differential equation this is reducible to linear okay but nollie is not not a linear differential equation but by some substitution we can say reduce this to linear differential equation and the last one is exact differential equation so we will talk about each one of these one by one so our first one is variables a variable method now here your differential equation will be in this form dy by DX this equals to some function of X multiplied by some function of Y okay if your differential equation is in this form then you can solve this differential equation using variable separable method so here the approach will be you have to write the function of X with DX and function of Y with dy so you can write this relation as dy by Z Y and this will be FX DX now you have to integrate a both side to get the solution of this differential equation so suppose if the integration of this is capital g of hawaii and this is capital f of X plus some integration constant C so this will be the solution of this differential equation now let me explain this idea with the help of one example suppose if you have this differential equation say dy by DX this equals to log X plus 1 by sine y plus cos y if you have this differential equation and you want to solve this differential equation then here you can easily see that you can write this function of Y with this dy and function of X with DX so you can write this relation as a sine y plus cos Y dy and this will be log X plus 1 DX now once you have separated the variable once you have written the function of Y with dy and function of X with DX then you can integrate a both sides to get the solution so if you integrate the both side then integration of sine Y will be minus Cosway integration of cos Y will be plus the sine Y here you have to integrate log X and integration one off one will be hits plus integration constant C okay first we have to integrate log X integration of one will be X plus C now we have to see how to integrate this log X so in order to integrate log its we have to use integration by parts formula we have seen what is integration by parts suppose if you have integration you be DX then this can be written as u integration of a DX minus integration of now you have to difference here this outer function so d u by DX integral of pdx and then you have to integrate this entire function so this formula is called u into V formula or integration by part now how to decide this U and V so u v is decided by this pilot rule where I stands for inverse function L for logarithmic a for algebraic t for trigonometric and E for exponential function so in this order the first function suppose if you have inverse and algebraic function so this means inverse will be your you and algebraic function will be function B so similarly here if I want to integrate this log X DX so this I can write as integration log X into 1 DX so now and you can see that this log X is logarithmic function 1 is algebraic function so here you can see that L is before a so this means you have to treat create this log X as you and one as a P so we can write this integral as log X integration 1 a DX minus derivative of log X 1 by X integration of 1 X DX so now this will give us from here you will get X so X log X minus these two will cancel so this will be the integration of log X integration of log X will be X log X minus X so finally your solution will be the solution will be your left-hand side is sine Y minus cos Y integration of log X is X log X minus X plus X plus this integration constant see these two will cancel and your final solution will be sine Y minus cos Y this equals to X log X plus C so this will be the solution of given differential equation now whenever you can separate the variable whenever you have some function of X multiplied or divided by some function of Z then in that case you can easily separate the variable and then by integrating both sides you can find the solution of the given differential equation so you can copy this then we'll take some more examples on this variable separable method now we have these two problems from gate previous here paper based on the concept of variable separable method the first question is from GAE 2015 in AC department the second one was asking engineering services 2017 so we are going to solve this question one by one the first problem is the given differential equation is dy by DX this equals to 1 plus cos Y but 1 plus cos 2 y by 1 minus cos 2x one more thing here the options are option a with minus sign option B with plus ly options he with minus sign ups and D with plus thing so first step will be here we need to separate the variables so dy by 1 plus cos 2 y and this will be equal to DX by 1 minus cos 2x now once we have written the function of x width DX and a function of y with dy we need to integrate both side to get the solution now if you want to integrate this 1 plus cos 2 y this will be 2 cos square Y 1 minus cos 2 X this will be 2 sine square X 2 and the 2 will can one by cos square Y will be six square ydy and here they will get Kousaka square X DX now integration of Seca square Y this we all know is nothing but tan Y and right hand side this will be minus cot X plus C if we rearrange this relation then it will get a tan y plus cot x equals to C so this will be our option B option Hays 10y minus cot x equals to C option B is tan y plus cot x equals to C so here the correct answer will be option B tan y plus cot x equals to C now the second problem that we have from this topic is we need to find the solution of this differential equation the given differential equation is y into root over 1 minus X square dy is minus X into root over 1 minus y square DX so here again you can easily separate the variables variables and once you have separated the variable then you have to integrate both sides to get the solution so if we separate the variable then you can write Y dy divided by root over 1 minus value square and this will be minus X DX divided by root over 1 minus X square now you have to integrate both side so these two integrals are similar if we take this minus sign outside and then you can see that here you have Y by root over 1 minus y square here you have X by root over 1 minus X square so let me call one of these integrals as i1 okay let me call this integral as Phi so I guess suppose x DX by we can solve this integral here so let let I be DX by let I is integration X DX by root over 1 minus X square now in order to solve this integral we have to do this substitution for 2 1 minus X square equals 2t so this means a DT will be minus 2 X DX which means in place of X DX you can write minus DT by 2 so this integral will become integration minus the DT by 2 divided by 1 minus X s square s T so this will become root 2 T so minus 1 by 2 an integral of 1 by Rho T this will be nothing but 2 root 2 T 2 n 2 will cancel finally you will get a minus and a rota T is nothing but root over 1 minus X square so this is the value of this integral integration xdx by 1 minus X square root over 1 minus xn Square this will be minus root over 1 minus X square so now you can see what should be the value of this first integral this will be nothing but minus root over 1 minus X Y Square and from here you will get minus root over 1 minus X square plus is C okay so now if you want to rearrange this relation you can write this as minus minus plus so root over 1 minus X s square here you will get root over 1 minus y square plus integration constant C equals to 0 so this will be the correct answer now you can see that this is nothing but option C okay whether you write let's see or minus C this will not make any difference because C is only arbitrary constant so here the correct answer will be option C now after this we will talk about reducible to variable separable differential equations the differential equation you know in those type of problem you cannot directly separate the variable but by doing some substitution you can separate the variable and then we will follow the same approach okay so next we have to learn how to solve a differential equation which are reducible to variable separable so here the standard format will be dy by DX this equals to some function of ax plus b y plus a C okay I remember that in these type of differential equation first to see whether you can directly separate the variables or not if you can directly separate the variable then separate the variable and integrate on both sides to get the solution but if you cannot separate the variable then you have to perform this substitution ax + B Y plus C this equals 2t okay in place of X plus B Y plus C you have to write a T and then you will see that you can separate the variable so for example suppose if I have to solve this differential equation say dy by DX this equals 2 cosine of X plus y say I had this equation so here if I want to separate the variable then you cannot directly separate the variable because you have cos and X plus y is inside this cosine function so here in order to separate the variable you have to do this substitution X plus y equals 2t because you have this X plus y inside cosine now if you write X plus y as T then you also have to replace this dy by DX for that you have to use this relation so if I difference here this is a lesson with respect to I X I will get a 1 plus dy by D X this will be DT by DX so this means a dy by DX will be DT by DX minus 1 so now in a place of dy by DX I will write DT by DX minus 1 and in place of X plus y I will write P so our differential equation will become DT by DX minus 1 this equals 2 cosine a D so this means DT by DX this will be 1 plus 4 sine of T now you can see that you have this function of T here so you can separate these variables you can write the funk of tea with DT so DT by 1 plus cosine NT this will be DX once you have separated the variable you have to integrate both sides to get the solution so if I integrate this then of 1 plus cos T this will be 2 cos a square at ey u du and here you will get X plus C so now you will get 1 by 2 integration Seca Square t by 2 DT and right hand side will be X plus C 1 by 2 integration of Seck a square t by 2 this will be tan a t by 2 and in integration we have to divide by the coefficient of T this will be x + C 1 by 2 will cancel finally you will get a tan and in place of T now we have to substitute x + y because our original differential equation is in the form of x + y so I will get tan of x + y by 2 this equals 2x + C and this will be the solution of given differential equation so here you can see that you cannot directly separate the variable but once you have done this substitution X plus y equals 2t then you can easily separate the variables so we have one more problem based on this variable separable method so that will dismiss after this problem ok so the next problem that we have on the concept that we have learned just now is this question was asking a 20-17 the problem is the solution of this differential equation dy by DX is X plus y -1 squared is 4 options are given option a is with plus sign option B is with minus sign option C with plus sign and B with minus sign so we have to find which one of these function is the solution of this given differential equation now again as you can see you cannot separate the variables here directly so the given differential equation here is dy by DX this equals 2x plus y -1 now since you cannot separate the variables directly in place of X plus y minus one you have to write a t because now this right hand side is function of X plus y minus one so if I write x plus y minus 1 equals 2t then we also need to replace dy by DX for that if I differentiate this relation I will get 1 plus dy by DX will be DT by DX so this means that dy by DX will be DT by DX minus 1 in place of dy by DX we can write DT by DX minus 1 so now our differential equation will be this is your equation 1 so now you can write this differential equation 1 as in place of dy by DX we need to write a DT by DX minus 1 so DT by DX minus 1 and our right hand side will be X plus y minus 1 is nothing but 2 T so this will become T Square from here you can write DT by DX this will be T square plus 1 DT by T square plus 1 this will be DX once you have separated the variable you can integrate both side so integration of this will be tan inverse T integration of X will be X plus C so this means that T will be nothing but tan of X plus C T was nothing but X plus y minus 1 so this will be tan of X plus C so finally our Y will be 1 minus X plus tan of X plus C so now you can see that absent a and B are with tan inverse X plus C so this is incorrect option C is with plus sign here option D is with minus sign so you can see that Y is 1 minus X plus a tan of X plus C so this minus sign is nothing but option D so for this question the correct answer is option D now after this we will talk about homogeneous differential