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Solve the initial value problem:
dy
——— = 2xy², y = 2, when x = – 1.
dx
Separate the variables in the equation above:
[tex]\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\ \mathsf{y^{-2}\,dy=2x\,dx}[/tex]
Integrate both sides:
[tex]\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}[/tex]
[tex]\mathsf{\dfrac{1}{y}=-(x^2+C_1)}[/tex]
Take the reciprocal of both sides, and then you have
[tex]\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}[/tex]
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,
[tex]\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}[/tex]
[tex]\mathsf{C_1=-\,\dfrac{3}{2}}[/tex]
Substitute that for C₁ into (i), and you have
[tex]\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}}[/tex]
So y(– 2) is
[tex]\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.}[/tex]
I hope this helps. =)
Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus
—————
Solve the initial value problem:
dy
——— = 2xy², y = 2, when x = – 1.
dx
Separate the variables in the equation above:
[tex]\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\ \mathsf{y^{-2}\,dy=2x\,dx}[/tex]
Integrate both sides:
[tex]\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}[/tex]
[tex]\mathsf{\dfrac{1}{y}=-(x^2+C_1)}[/tex]
Take the reciprocal of both sides, and then you have
[tex]\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}[/tex]
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,
[tex]\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}[/tex]
[tex]\mathsf{C_1=-\,\dfrac{3}{2}}[/tex]
Substitute that for C₁ into (i), and you have
[tex]\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}}[/tex]
So y(– 2) is
[tex]\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.}[/tex]
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:
[tex]\frac{dy}{dx} = 2xy^2[/tex]
What is separation of variables?
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:
[tex]\frac{dy}{dx} = 2xy^2[/tex]
[tex]\frac{dy}{y^2} = 2x dx[/tex]
[tex]\int y^{-2} dy = \int 2x dx[/tex]
[tex]-\frac{1}{y} = x^2 + C[/tex]
[tex]y(x^2 + C) = -1[/tex]
[tex]y = -\frac{1}{x^2 + C}[/tex]
y(-1) = 2, hence when [tex]x = -1, y = 2[/tex], and this is used to find C.
[tex]2 = -\frac{1}{(-1)^2 + C}[/tex]
[tex]2 + 2C = -1[/tex]
[tex]2C = -3[/tex]
[tex]C = -\frac{3}{2}[/tex]
Hence the solution is:
[tex]y = -\frac{1}{x^2 - \frac{3}{2}}[/tex]
When x = 2, we have that:
[tex]y = -\frac{1}{2^2 - \frac{3}{2}} = -\frac{1}{\frac{1}{2}} = -2[/tex]
Thus, y(2) = -2.
To learn more about separation of variables, you can take a look at https://brainly.com/question/14318343
