Respuesta :
Answer:
The solution is [tex]y=-\frac{12}{12x-30x^2+1}[/tex].
Step-by-step explanation:
A first order differential equation [tex]y'=f(x,y)[/tex] is called a separable equation if the function [tex]f(x,y)[/tex] can be factored into the product of two functions of [tex]x[/tex] and [tex]y[/tex]:
[tex]f(x,y)=p(x)h(y)[/tex]
where [tex]p(x)[/tex] and [tex]h(y)[/tex] are continuous functions.
We have the following differential equation
[tex]y'=(1-5x)y^2, \quad y(0)=-12[/tex]
In the given case [tex]p(x)=1-5x[/tex] and [tex]h(y)=y^2[/tex].
We divide the equation by [tex]h(y)[/tex] and move [tex]dx[/tex] to the right side:
[tex]\frac{1}{y^2}dy\:=(1-5x)dx[/tex]
Next, integrate both sides:
[tex]\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C[/tex]
Now, we solve for [tex]y[/tex]
[tex]-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}[/tex]
We use the initial condition [tex]y(0)=-12[/tex] to find the value of C.
[tex]-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}[/tex]
Therefore,
[tex]y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}[/tex]
