- #1

dagg3r

- 67

- 0

and answers but not sure if they are right mind if someone

can check them for me thanks

Find the general solution of the differential equation

dy/dx - 2y = e^(5x)

i found I(x) = e^ integral (-2 dx) = e^(-2x)

as I(x) = e^Integral (p(x)) in this case p=-2

therfore d/dx((e^-2x)*y) = e^(3x) // as we multiply both sides by e^(-2x)

therefore e^(-2x)y = integral e^(3x)

e^(-2x)y = e^(3x)/3 + C

y= e^(3x)/(3*e^(-2x)) + C

y= 1/3e^(5x) + Ce^(2x)

2.

given that z=2xt^3 - cos5x and x^3 + t^2 = 6x

find dz/dt using a chain rule

i will use notatio 9 as the "day symbol"

dz/dt = 9z/9x*dx/dt + dz/dt

= (2t^3 + 5sin5x)*dx/dt + 6xt^2 !(3)!

then knowing

x^3 + t^2 = 6x i diff resp to x so

3x^2*dx/dt - 6dx/dt = -2t

dx/dt = 2t/(3x^2 - 6)

i subtituted this back into (3) and got

= (4t^4 + 10tsin5x + 18x^3t^2 - 36xt^2) / (3x^2-6)

3. given that z=e^(5x)sin2y and y=x^2 + 5

find dz/dx as a function of x using a chain rule

same 9 symbol represtn day

dz/dx=9z/9x + 9z/9y*dy\dx

dz/dx = 5e^(5x)sin2y + 2xe^(5x)cos2y

i subtitued y=x^2 + 5 and got this

dx/dx = e^(5x) * [5sin(2x^2+10) + 2xcos(2x^2 + 10) ]

so i hope i did it right

thanks for the help guys