How do you solve the differential equation $dy/dx=6y^2x$ , where $y(1)=1/25$ ?

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Answer 1 · 2018-04-26T19:09:11

$=> y = 1/(28-3x^2)$

We are given:

$dy/dx = 6y^2x$

Separate the variables:

$1/y^2 dy = 6x dx$

Integrate both sides:

$int 1/y^2 dy = int 6x dx$

$-1/y = 3x^2 + C$

$1/y = -3x^2+C$

$y = -1/(3x^2 + C)$

where $C$ is an arbitrary constant of integration.

Now solve for $y(1)$ to find $C$ :

$y(1) = 1/25 = -1/(3(1)^2+C)$

$-1/25 = 1/(3+C)$

$3+C = -25$

$C = -28$

Hence, the final solution is:

$y = -1/(3x^2-28)$

$=> y = 1/(28-3x^2)$

How do you solve the differential equation dy/dx=6y^2xdy/dx=6y^2x, where y(1)=1/25y(1)=1/25 ?