Question #31289

2 Answers
Mar 29, 2017

#y = 2/3 x^3 +x^2-2/3 x+ 2#

Explanation:

If #(d^2y)/dx^2 = 2-4x# then #(dy)/(dx)=2x-1/2xx 4 x^2+ a# and

#y = 1/2 xx 2 x^2-1/3 1/2 xx 4 x^3 + a x + b = x^2+2/3 x^3+a x+ b#

Now we know

#{((-1)^2+2/3(-1)^3+a(-1)+b = 3),(0^2+2/3 0^3+a cdot 0+b = 2):}#

Solving for #a,b# we have

#a=-2/3, b = 2# so the curve is

#y = 2/3 x^3 +x^2-2/3 x+ 2#

Mar 29, 2017

#y(x) = -2/3x^3+x^2+2/3x+2#

Explanation:

This is a separable differential equation, so we can find the general solution by integration:

#(d^2y)/(dx^2) = 2-4x#

#(dy)/(dx) = int (2-4x)dx = 2x-2x^2+C_1#

#y = int (2x-2x^2+C_1)dx = x^2-2/3x^3+C_1x+C_2#

We can now find the values of #C_1# and #C_2# from the equations we get from the initial conditions:

#y(-1) = 3#

#y(0) = 2#

so that:

#{(1+2/3-C_1+C_2 = 3),(C_2 = 2):}#

#{(C_1 = -3+2+1+2/3 = 2/3),(C_2 = 2):}#

and we can conclude that:

#y(x) = -2/3x^3+x^2+2/3x+2#