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

ecoanuncios7x
2022-09-27
Answered

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

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cisalislewagobc

Answered 2022-09-28
Author has **8** answers

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

Separate the variables:

$\frac{1}{{y}^{2}}dy=6xdx$

Integrate both sides:

$\int \frac{1}{{y}^{2}}dy=\int 6xdx$

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

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

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

where C is an arbitrary constant of integration.

Now solve for y(1) to find C:

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

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

$3+C=-25$

$C=-28$

Hence, the final solution is:

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

$\Rightarrow y=\frac{1}{28-3{x}^{2}}$

Separate the variables:

$\frac{1}{{y}^{2}}dy=6xdx$

Integrate both sides:

$\int \frac{1}{{y}^{2}}dy=\int 6xdx$

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

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

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

where C is an arbitrary constant of integration.

Now solve for y(1) to find C:

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

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

$3+C=-25$

$C=-28$

Hence, the final solution is:

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

$\Rightarrow y=\frac{1}{28-3{x}^{2}}$

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