Question

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Answer 1

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Solve the initial value problem:

   dy
———  =  2xy²,      y = 2,  when x = – 1.
   dx


Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\ \mathsf{y^{-2}\,dy=2x\,dx}$


Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\ \mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\ \mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\ \mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$


Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$


In order to find the value of  C₁  , just plug in the equation above those known values for  x  and  y, then solve it for  C₁:

y = 2,  when  x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\ \mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\ \mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\ \mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\ \mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$


Substitute that for  C₁  into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\ \mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\ \mathsf{y=-\,\dfrac{2}{2x^2-3}}$


So  y(– 2)  is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\ \mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$


I hope this helps. =)


Tags:  ordinary differential equation ode integration separable variables initial value problem differential integral calculus