Question

y' +y =2
y(0)=2

Find a solution using integrating factor.

Answer 1

The integrating factor would be e^x right?

Answer 2

2(y - 4x^2)dx + (x)dy = 0

2(y - 4x^2) + (x)(dy/dx) = 0

2y - 8x^2 + xy' = 0

xy' + 2y = 8x^2

y' + (2/x)y = 8x

I.F. = e^ int(2/x)dx
= e^(2lnx)
= e^(lnx^2)
= x^2

(x^2)y' + (x^2)(2/x)y = (x^2)8x

(x^2)y' + (2x)y = 8x^3

d((x^2)y)/dx = 8x^3

d((x^2)y) = 8x^3 dx

(x^2)y = 2x^4 + C

Answer 3

IF = e^(INT1dx) = e^x

right

Answer 4

What would my next step be @cwrw238 ?

Answer 5

thanks @abayomi12 , it might be better if you explained but you did help me with another question i had so thumbs up for you (not literally)

Answer 6

next step s to multiply each term by e^x

Answer 7

\[e ^{x}y' +e ^{x}y = 2e^{x}\]

Answer 8

ok - note that left hand side is the result of differentiating ye^x so:

ye^x = INT 2e^x

Answer 9

\[y=e ^{-x}(2e ^{x}+C)\]

Answer 10

\[y=2+Ce ^{-x}\]
is it -x or x?

Answer 11

so y=2, awesome thanks @cwrw238

Answer 12

-x

Answer 13

yw