What is the general solution for the differential equation? #(dy)/(dx) = (3x^2 + 3cosx + 4e^3x)(y^2 + 4)#

1 Answer
Truong-Son N.
Jun 14, 2017

#y_g = 2tan(2x^3 + 6sinx + 4e^3x^2 + D)#

#npi + pi/2 < 2x^3 + 6sinx + 4e^3x^2 + D < npi + (3pi)/2#,

#n in ZZ#

For example, if #D = 1#, then the graph looks like this:

Move your #x# terms to one side and #y# terms to the other.

#1/(y^2 + 4)dy = (3x^2 + 3cosx + 4e^3x)dx#

Integrate to get:

#int 1/(y^2 + 4)dy = int (3x^2 + 3cosx + 4e^3x)dx#

#1/4 int 1/((y/2)^2 + 1)dy = int (3x^2 + 3cosx + 4e^3x)dx#

#1/2 arctan(y/2) = x^3 + 3sinx + 2e^3x^2 + C#

Solving for #y# multiply by #2#, take the #tan# of both sides, and multiply by #2# again.

We get a (very ugly) general solution of:

#color(blue)(y_g = 2tan(2x^3 + 6sinx + 4e^3x^2 + D))#,

#npi + pi/2 < 2x^3 + 6sinx + 4e^3x^2 + D < npi + (3pi)/2#,

#n in ZZ#,

where #D = 2C# is a new arbitrary constant.

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