What's the particular solution that satisfies #y(0)=0# of #dy/dx=(-2x+y)^2-7# ?

1 Answer
Apr 17, 2018

# y = (2z+3 -2xe^(6x) + 3e^(6x))/(1-e^(6x))#

Explanation:

We have:

#dy/dx=(-2x+y)^2-7# with #y(0)=0#

We can perform a substitution:

Let# u=-2x+y \ # and # \ x=0,y=0 => u=0#

If we differentiate wrt #x#, then:

# (du)/dx = -2+dy/dx => dy/dx = (du)/dx +2#

Substituting into the initial DE:

# (du)/(dx) + 2 = u^2-7# with #u(0)=0#

# :. (du)/(dx) = u^2-9#

# :. 1/(u^2-9) (du)/(dx) = 1 #

This has reduced the equation to a First Order Separable ODE, so we can "separate the variable":

# int \ 1/(u^2-9) \ du =int \ u \ du #

To integrate the LHS we can decompose the integrand into partial fractions:

# 1/(u^2-9) -= 1/((u-3)(u+3)) -= A/(u-3) + B/(u+3) #

Using the "cover-up" method we then have:

# 1/((u-3)(u+3)) -= (1/6)/(u-3) + (-1/6)/(u+3) #

So then we have:

# 1/6 int \ 1/(u-3) - 1/(u+3) \ du = int \ dx#

Which we can now readily integrate:

# 1/6 {ln|u-3| - ln |u+3|} = x + C#

Applying the initial condition #u(0)=0#

# 1/6 {ln|-3| - ln |3|} = C => C = 0#

So we get a Particular Solution:

# ln|u-3| - ln |u+3| = 6x #

# :. ln|(u-3)/(u+3)| = 6x #

# :. |(u-3)/(u+3)| = e^(6x) #

Noting that #e^a gt 0 AA a in RR#

# (u-3)/(u+3) = e^(6x) #

# :. u-3 = ue^(6x) +3 e^(6x)#

Restoring the earlier substitution #u=-2x+y#

# -2x+y-3 = (-2x+y)e^(6x) +3e^(6x)#

# :. -2x+y-3 = -2xe^(6x) +ye^(6x) + 3e^(6x)#

# :. y - ye^(6x) = 2z+3 -2xe^(6x) + 3e^(6x)#

# :. y = (2z+3 -2xe^(6x) + 3e^(6x))/(1-e^(6x))#