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

Question

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).

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Eloise 5 days 2022-01-08T20:50:22+00:00 1 Answer 0 views 0

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    2022-01-08T20:51:50+00:00

    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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