What is the general solution of the differential equation $\frac{d y}{d x} + y^{2} = 4$ $dy/dx + y^2=4$ ?

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Answer 1 · 2017-09-16T23:37:53

$y = 2 \frac{e^{4 x} - 16}{e^{4 x} + 16}$ $y = 2(e^(4x)-16)/(e^(4x)+16)$

We have:

$\frac{d y}{d x} + y^{2} = 4$ $dy/dx + y^2=4$

We can rearrange this Differential Equation as follows:

$\frac{d y}{d x} = 4 - y^{2}$ $dy/dx =4 -y^2$
$:. 1/(4-y^2) dy/dx = 1$

This is a First Order separable Differential Equation, so we can "seperate the variables" to get:

$int \ 1/( 2^2-y^2) \ dy = int \ dx$ ..... [A]

Using the standard result:

$int \ 1/(a^2-t^2) \ dt = 1/(2a) ln abs( (a+t)/(a-t) )$

we can integrate [A], to get:

$1/4 ln abs( (2+y)/(2-y) ) = x + C_1$
$:. ln abs( (2+y)/(2-y) ) = 4x + C$

Using the initial condition, $x=ln2$ when $y=0$ then:

$-ln abs( (2+0)/(2-0) ) = 4ln2 + C$
$:. 0 = 4ln2 + C$
$:. C = -4ln2 = - ln16$

Thus, the (implicit) solution is:

$ln abs( (2+y)/(2-y) ) = 4x - ln16$

We can rearrange to get an explicit solution:

$:. abs( (2+y)/(2-y) ) = e^(4x - ln16)$

Noting that $e^x gt 0 AA x in RR$ , then:

$(2+y)/(2-y) = e^(4x)e^(-ln16)$
$:. (2+y)/(2-y) = e^(4x)/16$

For simplicity, Put $A=e^(4x)/16$ so that:

$(2+y)/(2-y) = A$
$:. 2+y = A(2-y)$
$:. 2+y = 2A-Ay$
$:. y+Ay = 2A-2$
$:. y(A+1) = 2(A-1)$
$:. y = 2(A-1)/(A+1)$

Thus, the solution is:

$y = 2(e^(4x)/16-1)/(e^(4x)/16+1)$
$\ \ = 2(e^(4x)/16-1)/(e^(4x)/16+1) xx 16/16$
$\ \ = 2(e^(4x)-16)/(e^(4x)+16)$

What is the general solution of the differential equation dy/dx + y^2=4dydx+y2=4 dy/dx + y^2=4?