00:01
We will now determine the solution to this given differential equation, which is 2nd squared y, dua by dx plus 2x tan y equal to x cube, is the differential equation.
00:12
So how do we solve this one? we first make a substitution which will convert this into a linear differential equation of first order, and then we follow the steps to find solution of any linear differential equations of first order.
00:28
So basically we have two steps that we should follow to determine the solution of a linear differential equation.
00:36
If the equation is in the form of d .y by dx plus py equal to q, where p and q are functions of x, we first find the integrating factor, which is given as e raised to the power of integrated pdx.
00:49
And then we find the solution as y times of integrating factor is equal to integration of q terms of integrating factor dx.
00:58
Plus c where c is the constant of integration so let's go ahead and do the substitution which is our first task to convert this into a linear differential equation as you can notice we have this sequence squared y which is basically the differentiation of tan y i'm going to make this substitution tan y equal to v and now i differentiate both sides with respect to x which means this will become second squared y d y by d x equal to dv by d x observe that we already have this term second squared y d y by d x in this given differential equation which we will now replace this as dv by d x so at forth the the given equation will become dv by d x plus 2x tan y is basically v this is equal to x cube so this is now a linear differential equation of first order in terms of we so we will find the solution by following the steps the first thing is that we have to recognize the values of p and q so here the value of p equal to except the v which is 2x and then the value of q is equal to the term on the right side that is x cube so let's find the integrating factor which is e raised to the power of p d x this is equal to e raised to the power of substitute for p which is 2x d x notice that integration of 2x dx is basically when you apply the power rule of integration we'll get x squared by 2 so that it just becomes x squared so this is basically e raised to the power of x squared this is our integrating factor we then find the solution equation we have to trace this second rule which states that y times of integrating factor so i put y times of i know that integrating factor is e raised to the power of x quite remember that since we deal with the equation in terms of v here we have to put v times of e power x squared so this is equal to integration of q times of q is basically x q times of integrating factor which is a e raised to the power of x squared x squared dx plus the constant of integration which is c we now solve this equation for we first let's differentiate this part so to differentiate this part we can make a temporary substitution that is a let's put x squared equal to t we then differentiate both sides with respect to x and with respect to t on the right side so this will give us 2x times of dx is equal to d t which means i can now solve for x dx so this gives me x dx is equal to d t by 2 so now since i'm going to make this substitution into the integration i'm going to make some adjustment i'm going to rewrite this x -cube as like this that is i can write down this as x squared then i put this e raised to the power of x squared.
04:32
I still have x which i left out for x cube.
04:36
So that i'm going to put it as xdx integration plus c.
04:43
So as you can observe that, i can replace this xdx using this substitution which is dt by 2.
04:50
So that this will now become, we already meet the substitution x squared equal to t.
04:56
So therefore this is equal to t.
04:57
E raised to the power of t and x dx is basically d t by two i put this as d t and put the half before the integration as like this plus c now to integrate this t t times of e raise to the power of t we have to use the integration by parts method integration by parts which i will do now we have to basically follow the rule ilate which stands for inverse logarithmic algebra trigon and exponential this rule we follow to select the u so if you can look at this product of two function t x times of e raise to the power of t the algebra function t comes in first so i can use the u as t and rest of them as dv so when i do that dv is equal to e power t d t d t now i have to differentiate this term that is a d u is equal to d t and here i have to integrate.
06:00
When i integrate here, i will get to v.
06:02
That is, integration of dv is like this.
06:05
This is equal to v.
06:07
Integration of e power t is e power t...