dy/dx = 2xy^2 and y(-1) = 2 find y(2)
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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
Using separation of variables, we have that y(2) = -2.
The differential equation is:
\(\frac{dy}{dx} = 2xy^2\)
In separation of variables, we place all the factors of y on one side of the equation with dy, all the factors of x on the other side with dx, and integrate both sides.
Hence:
\(\frac{dy}{dx} = 2xy^2\)
\(\frac{dy}{y^2} = 2x dx\)
\(\int y^{-2} dy = \int 2x dx\)
\(-\frac{1}{y} = x^2 + C\)
\(y(x^2 + C) = -1\)
\(y = -\frac{1}{x^2 + C}\)
y(-1) = 2, hence when \(x = -1, y = 2\), and this is used to find C.
\(2 = -\frac{1}{(-1)^2 + C}\)
\(2 + 2C = -1\)
\(2C = -3\)
\(C = -\frac{3}{2}\)
Hence the solution is:
\(y = -\frac{1}{x^2 - \frac{3}{2}}\)
When x = 2, we have that:
\(y = -\frac{1}{2^2 - \frac{3}{2}} = -\frac{1}{\frac{1}{2}} = -2\)
Thus, y(2) = -2.
To learn more about separation of variables, you can take a look at https://brainly.com/question/14318343
