Question

Find the solution the each of the following first order linear differential equations:

a) xy' -4y = 2 x^6

b) y' - 5y = 4e^7x

c) dy/dx + 2y = 2/(1+e^4x)

d) 1/2 di/dt + i = 4cos(3t)

Answer 1

a.
`y=(2)/(3)x^7+cx^4`

b.
`y=2e^(7x)-ce^(5x)`

c.
`y=e^(-2x)arctan(e^(2x))+ce^(-2x)`

d.
`i=e^(-2t)\left((8\left(3e^(2t)\sin \left(3t\right)+2e^(2t)\cos \left(3t\right)\right))/(13)+C\right)`

Explanation:

a) xy' -4y = 2 x^6

`xy'-4y=2x^6\\y'-(4)/(x)y=2x^5\\p(x)=(-4)/(x)\\Q(x)=2x^5\\\mu(x)=\int P(x)dx=\int (-4)/(x)dx=Ln|x|^(-4)\\y=e^(-\mu(x))\int {e^(\mu(x))Q(x)dx}\\y=x^4 \int {x^(-4)2x^6}dx\\y=(2)/(3)x^7+cx^4`

b) y' - 5y = 4e^7x

`y'-5y=4e^(7x)\\p(x)=-5\\Q(x)=4e^(7x)\\\mu(x)=\int P(x)dx=\int-5dx=-5x\\y=e^(-\mu(x))\int {e^(\mu(x))Q(x)dx}\\y=e^(5x)\int {e^(-5x)4e^(7x)}dx\\y=2e^(7x)-ce^(5x)`

c) dy/dx + 2y = 2/(1+e^4x)

`(dy)/(dx)+2y=(2)/(1+e^(4x))\\p(x)=2\\Q(x)=(2)/(1+e^(4x))\\\mu(x)=\int P(x)dx=\int 2dx=2x\\y=e^(-\mu(x))\int {e^(\mu(x))Q(x)dx}\\y=e^(-2x)\int {e^(2x)(2)/(1+e^(4x))}dx\\y=e^(-2x)arctan(e^(2x))+ce^(-2x)`

d) 1/2 di/dt + i = 4cos(3t)

`(1)/(2)(di)/(dt)+i=4cos(3t)\\(di)/(dt)+2i=8cos(3t)\\p(t)=2\\Q(t)=8cos(3t)\\\mu(t)=\int P(t)dt=\int 2dt=2t\\i=e^(-\mu(t))\int {e^(\mu(t))Q(t)dt}\\i=e^(-2t)\int {e^(2t)8cos(3t}dt\\i=e^(-2t)\left((8\left(3e^(2t)\sin \left(3t\right)+2e^(2t)\cos \left(3t\right)\right))/(13)+C\right)`