Differential Equations and Fourier Series

[NaN089]

Homework Statement

Q1) (dy/dx)= 2x(y²+9); y(0)=0
Q2) (x⁴+y²) dx - xy dy =0; y(2)=1
Q3) (dy/dt)= 4y+t
Q4) y"+2y'+y=0; y(0)=4 and y'(0)=-6
Q5) y"+3y'+2y= 30e^2t

3. Solutions found

A1) y= tan (x²/3)
A2) this is not an exact differential and so cannot be solved
A3) couldn't solve this yet
A4) y= 4e^-x -2xe^-x or y= 2e^-x(2-x)
A5) y= Ae^-x + Be^-2x + 2.5e^2t

Can someone check if my answers are correct, please?

Many Thanks

 

[Mark44]

Homework Statement

Q1) (dy/dx)= 2x(y²+9); y(0)=0
Q2) (x⁴+y²) dx - xy dy =0; y(2)=1
Q3) (dy/dt)= 4y+t
Q4) y"+2y'+y=0; y(0)=4 and y'(0)=-6
Q5) y"+3y'+2y= 30e^2t

3. Solutions found

A1) y= tan (x²/3)
A2) this is not an exact differential and so cannot be solved
A3) couldn't solve this yet
A4) y= 4e^-x -2xe^-x or y= 2e^-x(2-x)
A5) y= Ae^-x + Be^-2x + 2.5e^2t

Can someone check if my answers are correct, please?

For 1, 4, and 5, you should check your own work. For example, is tan(0) = 0? Do this function and its derivative satisfy the differential equation? If so, your solution is correct.

For 3, rewrite the equation as y' - 4y = t. There are several approaches you can take, one of which is to find an integrating factor to multiply both sides of the equation by. By eyeball, e^-4t appears to be the integrating factor to use here.

BTW, your title is misleading. These differential equations don't have anything to do with Fourier Series.

 

[Mark44]

For 2, make the substitution u = y/x or y = ux. From this you get y' = u'x + u. Substitute for y and y' in your differential equation to get an equation that is separable.

 

[NaN089]

Thanks for your help mark. i corrected the silly mistake that i made in 1, and cracked 2 and 3 with your hints.
I'm new in this forum and this is my first post. I have not learned all its features but i will soon discover how to use those symbols. ;p

Thanks again for your help. :D