Question

y'+y=x y(0)=3
linear equation
differential equation​

Answer 1

Answer:

Please verify my answer to the following differential equation:

y′′−xy′+y=0

Let y=∑∞n=0Cnxn, then y′=∑∞n=1nCnxn−1 and y′′=∑∞n=2n(n−1)Cnxn−2

Substituting this to the equation we get

∑n=2∞n(n−1)Cnxn−2−x∑n=1∞nCnxn−1+∑n=0∞Cnx

Answer 2

Working Rule :-

Let us consider a linear differential equation

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \: where \: p \: and \: q \in \: f(x)$

Step :- 1 :- Find Integrating Factor, I. F.

$\rm :\longmapsto\:I. F. = {e}^{ \: \int p \: dx}$

Step :- 2, Solution of differential equation is

$\rm :\longmapsto\:y \times I. F. = \int \: (q \times I. F.)dx$

Formula Used :-

$\boxed{ \red{ \bf \:\: \displaystyle \int \sf \: 1dx = x + c}}$

$\boxed{ \red{ \bf \:\: \displaystyle \int \sf \:u.v \: dx =u\displaystyle \int \sf \:vdx - \displaystyle \int \sf \:\bigg(\dfrac{d}{dx}u \:\displaystyle \int \sf \:vdx\bigg)dx}}$

Solution :-

Given differential equation is

$\rm :\longmapsto\:y' + y = x$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} + y = x$

On comparing with

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \:$

we get

$\rm :\longmapsto\:\boxed{ \red{ \bf \:p = 1}} \\ \rm :\longmapsto\:\boxed{ \red{ \bf \:q = x}}$

Now,

$\rm :\longmapsto\:I. F. = {e}^{ \: \int p \: dx}$

$\rm :\longmapsto\:I. F. = {e}^{ \: \int 1 \: dx}$

$\rm :\longmapsto\:I. F. = {e}^{ x}$

Hence,

Solution is given by

$\rm :\longmapsto\:y \times I. F. = \int \: (q \times I. F.)dx$

$\rm :\longmapsto\:y \times {e}^{x} = \int \: (x {e}^{x} )dx$

$\rm :\longmapsto\:y \times {e}^{x} = x\displaystyle \int \sf \:\: {e}^{x}dx \: - \: \displaystyle \int \sf \:\bigg(\dfrac{d}{dx}x\displaystyle \int \sf \: {e}^{x}dx\bigg)dx$

$\rm :\longmapsto\:y {e}^{x} = x {e}^{x} - {e}^{x} + c$

$\bf\implies \:y = x - 1 + c {e}^{ - x} - - (1)$

Now put x = 0 and y = 3, we get

$\bf\implies \:3 = 0 - 1 + c {e}^{ - 0}$

$\bf\implies \:c = 4$

Hence, equation (1) can be rewritten as

$\purple{\bf\implies \:y = x - 1 + 4 {e}^{ - x}}$