Checking solutions for first-order linear and chain rule problems

[dagg3r]

hi guys need some help on diff eqn, I've done the workingout
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

 

[HallsofIvy]

hi guys need some help on diff eqn, I've done the workingout
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)

Yes, that's correct.

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)

I think that's right but my feeling is you would be better off leaving it as 6xt^2-(4t^4+ 5sin5x)/(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) ]

You are missing a factor of 2: The derivative of sin 2y wrt y is 2 cos2y and the derivative of x^2+ 5 is 2x: that should be 4x cos(2x^2+ 10).

so i hope i did it right
thanks for the help guys

Was there a reason for titling this thread "second order differential equations"? There was only one differential equation and it was first order.