Equation of a curve using a given point and slope

[metalscot]

Find a curve whose slope at any point (x,y) is equal to 5y, and which passes through the point (1,-2)



All previous examples which i have seen of this type of question give a much more complex slope and always in terms of x.
This is causing me a bit of confusion with my answer. Should I work though it by integrating the slope and putting this equal to x instead of y i.e

x=5/2y^2 + C

then plug in my given x and y to obtain C=-9
giving a final equation of

x= 5/2y^2 -9

or should I work through using a general solution using arbitrary constant?

All help will be very much appreciated

 

[Mark44]

Find a curve whose slope at any point (x,y) is equal to 5y, and which passes through the point (1,-2)



All previous examples which i have seen of this type of question give a much more complex slope and always in terms of x.
This is causing me a bit of confusion with my answer. Should I work though it by integrating the slope and putting this equal to x instead of y i.e

x=5/2y^2 + C

then plug in my given x and y to obtain C=-9
giving a final equation of

x= 5/2y^2 -9

or should I work through using a general solution using arbitrary constant?

All help will be very much appreciated

This problem implicitly asks you to solve the initial condition differential equation, dy/dx = 5y, y(1) = -2.

Your solution doesn't work. For your equation, dy/dx = 1/(5y).

 

[metalscot]

This problem implicitly asks you to solve the initial condition differential equation, dy/dx = 5y, y(1) = -2.

Your solution doesn't work. For your equation, dy/dx = 1/(5y).

From the initail conditions given I then re-arrange to give dy/y = 5dx?

and integrate this to obtain lny = 5x +c

 

[HallsofIvy]

dy/dx= 5y so dy/y= 5dx has nothing to do with the initial condition, that y(1)= -2.

Be careful! Integrating dy/y does NOT give ln y. It gives ln|y|. Often, we are working with positive numbers, but the condition that y(1)= -2 makes the difference important! Now, what value of c gives y satisfying y(1)= -2?

 

[metalscot]

dy/dx= 5y so dy/y= 5dx has nothing to do with the initial condition, that y(1)= -2.

Be careful! Integrating dy/y does NOT give ln y. It gives ln|y|. Often, we are working with positive numbers, but the condition that y(1)= -2 makes the difference important! Now, what value of c gives y satisfying y(1)= -2?

Working through i get

y=Ae^5x

so A=-2e^-5

so y=-2e^-5e^5x ??

 

[Mark44]

Working through i get

y=Ae^5x

so A=-2e^-5

so y=-2e^-5e^5x ??

Sure, but you can check this yourself. Once you have found the solution, most of the hard work is done, and it's a simple matter to verify that your solution works.
Any solution you get must satisfy two conditions:
1) y' = 5y
2) y(1) = -2

If both of these are true for your solution, then your function is a solution to the initial condition problem.

Last, how you wrote your solution is not technically correct and it's more difficult to read than it needs to be.

y = -2e^(-5) * e^(5x)

or (using the X² button, in the expanded menu after you click Go Advanced), you can write it like this:
y = -2e^-5e^5x

or you can learn a bit of LaTeX to make it look like this:
[tex]y = -2e^{-5}e^{5x}[/tex]

 

[metalscot]

Sure, but you can check this yourself. Once you have found the solution, most of the hard work is done, and it's a simple matter to verify that your solution works.
Any solution you get must satisfy two conditions:
1) y' = 5y
2) y(1) = -2

If both of these are true for your solution, then your function is a solution to the initial condition problem.

[tex]y = -2e^{-5}e^{5x}[/tex]

Thanks a lot mark44 and halls of ivy. Appreciate your help