Feb 13 Homework solutions:

Answers:

a)  y=  +- sqrt (  x^2+2C )

b)    y = cuberoot of (x^3/3+ 2x +C)

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

d)   y = sqrt( 2 e^x + 14)

e)  y =  e^(x^2/2 + x + 0)

f) we will do in class

Complete solutions:

Solve for general solutions:

a)  dy/dx=x/y

seperate the variables (put all terms with x on one side)
  y dy= x dx

integrate both sides without bounds so we get a +C

   y^2/2 = x^2/2 +C

solve for y

   y=  +- sqrt (  x^2+2C )

note that there are many formulas for y which satisfy the same differential
equation since C can be any real number and there is a +-

check your answer by differentiating dy/dx

If you did not get this correct, study it and then try (b) and (c)
again before reading their solutions.


b)  dy/dx= (x^2+2)/ (3y^2)

    3 y^2 dy = (x^2+2) dx

    y^3  = x^3/3 + 2x +C

    y = cuberoot of (x^3/3+ 2x +C)

check your answer


c)  4yy' -3e^x=0

   4y (dy/dx) - 3 e^x =0

   4y (dy/dx) =3 e^x 

     4y dy =3 e^x dx 
   
     2y^2  = 3e^x +C

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

Check your answer


Solve for particular solutions:

d)  y y' -e^x =0       y(0)=4

start the same as with the general solution:

y (dy/dx) - e^x =0

y (dy/dx) = e^x 

y dy = e^x dx

y^2/2= e^x +C

y = +- sqrt( 2 e^x +2C)

check your ansewr in the ODE.

Now we find the particular solution which says y(0)=4
by plugging in x=0 and y=4 in our general solution and finding out
whether we should have a + or a - and what the value of C must be. 

4=  +- sqrt( 2 e^0 +2C)= +- sqrt( 2 +2C) 

since 4 is positive it must be a + not a minus

4=sqrt(2+2c)   16= 2 +2c  14=2c  7=c

so the particular solution is the general solution with a + and
c=7 as follows:

y = + sqrt( 2 e^x + 14)

check both the ode and the value y(0)=4.

If you did not get this problem correctly try (e) and (f)
again before just reading the solutions.

(e)  y(x+1)+y'=0   y(-2)=1

y(x+1) + dy/dx =0

y(x+1) dx + dy =0

(x+1) dx + (1/y) dy=0

(1/y) dy= -(x+1) dx

 Ln |y|  =  x^2/2 + x +C

|y| = e^(x^2/2 + x + C)

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

now check this general solution and then use y(-2)=1 so substitute
x=-2 and y=1 to solve for C

1=   + - e^((-2)^2/2 + (-2) + C)

so it is a + not a -

Ln(1)= (-2)^2/2 + (-2) + C

and Ln(1)=0 so

0= 4/2 -2 + C  gives C=0

so the particular solution is

y =  +  e^(x^2/2 + x + 0)


f)  y(x+1) +y'=0  where y(-2)=1

this one I'll do in class



Prof Sormani