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

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Answer 1 · 2018-04-26T19:09:11

$\Rightarrow y = \frac{1}{28 - 3 x^{2}}$ $=> y = 1/(28-3x^2)$

We are given:

$\frac{d y}{d x} = 6 y^{2} x$ $dy/dx = 6y^2x$

Separate the variables:

$\frac{1}{y^{2}} d y = 6 x d x$ $1/y^2 dy = 6x dx$

Integrate both sides:

$\int \frac{1}{y^{2}} d y = \int 6 x d x$ $int 1/y^2 dy = int 6x dx$

$- \frac{1}{y} = 3 x^{2} + C$ $-1/y = 3x^2 + C$

$\frac{1}{y} = - 3 x^{2} + C$ $1/y = -3x^2+C$

$y = - \frac{1}{3 x^{2} + C}$ $y = -1/(3x^2 + C)$

where $C$ $C$ is an arbitrary constant of integration.

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

$y (1) = \frac{1}{25} = - \frac{1}{3 {(1)}^{2} + C}$ $y(1) = 1/25 = -1/(3(1)^2+C)$

$- \frac{1}{25} = \frac{1}{3 + C}$ $-1/25 = 1/(3+C)$

$3 + C = - 25$ $3+C = -25$

$C = - 28$ $C = -28$

Hence, the final solution is:

$y = - \frac{1}{3 x^{2} - 28}$ $y = -1/(3x^2-28)$

$\Rightarrow y = \frac{1}{28 - 3 x^{2}}$ $=> y = 1/(28-3x^2)$

How do you solve the differential equation dy/dx=6y^2xdydx=6y2xdy/dx=6y^2x, where y(1)=1/25y(1)=125y(1)=1/25 ?