Question

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

Answer 1

Answer:

a. $y=\frac{2}{3}x^7+cx^4$

b. $y=2e^{7x}-ce^{5x}$

c. $y=e^{-2x}arctan(e^{2x})+ce^{-2x}$

d. $i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)$

Step-by-step explanation:

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

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

 

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)

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

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

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