(a) We have y(0)=c1+c2=0, y^''(1)=c1 e+c2 e^-1=1 so that c1=e /(e^2-1) and c2=-e /(e^2-1) . The

(a) We have y(0)=c1+c2=0, y^''(1)=c1 e+c2 e^-1=1 so that...

Key Concepts

Boundary Value Problems

Boundary value problems involve solving differential equations subject to given conditions at different points, rather than just initial conditions at a single point. In these problems, the conditions are specified at the endpoints of the interval of interest, which requires determining specific parameter values in the general solution to meet all constraints.

Linear Differential Equations with Constant Coefficients

Linear differential equations with constant coefficients have solutions that are structured as combinations of functions whose form is determined by the characteristic equation. These equations allow for superposition of solutions, and the use of exponential functions or hyperbolic functions arises naturally as general forms depending on the nature of the eigenvalues of the system.

Hyperbolic Functions

Hyperbolic functions, such as sinh and cosh, are analogues of the trigonometric functions but based on hyperbolas instead of circles. They provide an alternative way to represent solutions to differential equations, especially when the characteristic roots are real. Their connection with exponential functions (e.g., sinh x = (e^x - e^(-x))/2) is essential for transforming between different solution representations.

Equivalence of Different Functional Representations

The equivalence of different functional representations, such as the use of exponentials versus hyperbolic functions, plays a crucial role in understanding solutions to differential equations. Recognizing that expressions in terms of exponentials can be rewritten in terms of hyperbolic functions (and vice versa) allows a deeper insight into the properties and behaviors of the solutions under various boundary conditions.

Transcript

00:01 In this multiple correct answer type question, it is asking that we have to find the solution of given differential equation that is dy by d x plus x is equal to x times of e to the power of parentheses of n minus 1 times of y.

00:29 So there are four options givens.

00:33 So my answer is for given question i consider that is a and b.

00:39 Now i will explain the answer.

00:44 So explanation will be that is first write the given differential equation that is dy by dx plus x is equal to x times of e to the power of parenthesis of n minus 1 times of y.

01:01 Now next step is take a, sorry, subtract x to both sides and take a common x on the right side so it will be written as that is this differential equation will be written as in the next step that is i can see that d .y by dx will be equal to after doing this operation i can say that x times of bracket in the bracket the term will be written as that is a to the power of n minus 1 times of y minus 1 now next step is it will be written as in the earth other way that is i can see that d -y over e to the power of n minus 1 pantheism of n minus 1 times of y minus 1 will be equal to x d x so here you can see that i had separated the x variable and y variable so why you variable be written as in the left side and x variable will be written as on the right side so now next step is take a integral on the the both side and simplify it then right side will be written as that is integration of x d x will be equal to i consider x squared over 2 plus c and now left side be written as that is indication of d y over e to the power into the power of panaceous n minus 1 times of y minus 1 will be written as for getting this answer on the left side i will do some more steps separately.

03:01 So here i will consider that e to the power of parenthesis of n minus 1 times of y minus 1 is equal to u.

03:15 So it will be tinnous that is i can say that e to the power of parenthesis of n minus 1 minus times of y it will be equal to u plus 1.

03:30 Now take a differentiation on both side.

03:33 So it will be written as that is i consider that left side will be that is e to the power of n minus 1 times of y times of parenthesis of n minus 1 d .y is equal to d u so now i can say that this is a equation number 1 so dye will be written as that is from this step i consider it d u over pantheises of n minus 1 and times of e to the power of of n minus 1 times of y will be written as by using this step equation number 1.

04:15 I can see that that is u plus 1 parenthesis of u plus 1.

04:22 So i can say that this is an equation number 2.

04:31 Now in the above step the left side integral is, that is i will write this integral here, that is integral dy over e to the power of parenthesis of n minus 1 times of y minus 1 will be equal to by using this equation number 1 and equation number 2 it will be written as in the u variable i can see that d y will be replaced by this equation number 2 so i consider at d u over n minus 1 parenthesis of n minus 1 times of u plus 1 and e to the power of n minus 1 times of y minus 1 will be written by using this equation number 1 i can see that that is u so now 1 over n minus 1 is a constant value so i will take outside so in the integral inside inside will be 10 as it will be 10 as 1 over n minus 1 and inside will be 10 as it is integral d u over u plus 1 times of you now the integrant will be written as in the other way i can see that 1 over and minus 1 times of integral and numerator of integrant will be ten as that is u plus 1 minus u divided by parenthesis of u plus 1 times of u so here i had did some algebra manipulation in the numerator of integrant so i had taken two terms that is u plus 1 and u and it will be ten as is u plus 1 minus u so you can see that the numerator is as it is that is on the above step that is numerator is that is 1 so after subtracting u plus 1 sorry u from u plus 1 that will be due that is 1 so after doing this algebra manipulation again it will be written as that is in the other way i i consider that 1 over n minus 1 and integral in the bracket i will write that is after dividing by u plus 1 parenthesis of u plus 1 times of u with the every term of the numerator so it will be written as there is 1 over u minus 1 over u plus 1 d u now simplify it after taking a integration so it will be written as there is 1 over n minus 1 and integral of 1 over u du will be written as that is log u.

07:44 Base will be that is e and minus 1 over integral 1 over u plus 1 d u will be written as that is by using the standard indication formula i can say that that is log u plus 1 basis that is e.

08:04 So now next step is using the log property i can say that it will be written as that is 1 over n minus 1.

08:15 1 % % % % % x0x1 times of in the bracket after using this log property that is log a minus log b will be equal to log a over b so it will be written as that is log u over u plus 1.

08:36 Now next step is use this equation number 1 that is this equation and i will change this result in the by variable so it will be written as that is equal to 1 over n minus 1 times of in the bracket i will write that is log of basis that is e so u over u plus 1 will be written as that is e to the power of parenthesis of n minus 1 times of y minus 1 divided by e to the power of parenthesis of n minus 1 times of y so now next step is you can see that i had get the answer for integration of left side in this above step in this above previous step.

09:42 So i will substitute this value, this result for this integral which is written on the right side.

09:53 So the next step will be written as that is the left side.

09:58 Will be i consider the integral of du over e to the power of n minus 1 times of y minus 1 will be written as that is 1 over n minus 1 times of log of e to the power of parenthesis of n minus 1 times of y minus 1 divided by e to the power of parenthesis of n minus 1 times of y is equal to right side will be that is after taking this indication of x about x then i will get the answer on the right side that is x over 2 plus c so i will write this here in this step that is equal to x over 2 plus c so now look the option then you get the answer which is matching with the option i can say that this is the option a so now next step is that is multiply by n minus 1 to both sides...

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