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monitta
3 years ago
12

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

in parentheses. For example, (a−b)/(1+n).
Mathematics
1 answer:
Nata [24]3 years ago
4 0

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\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}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

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