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Solution of diffrential equation $6\frac{dy}{dx} -2y =xy^4$ is

Option 1) 2+(x+1)y³=2cy³e^x

Option 2) 2-(x+1)y³=2cy³e^-x

Option 3)2+(x-1)y³=2cy³e^-x

Option 4) 2-(x+1)y³=2cy³e^x

Answers (1)

best_answer

As we have learned

Bernoulli's Equation -

$\frac{1}{y^{n-1}}= v$

$\frac{1}{y^{n}}\frac{dy}{dx}= \frac{1}{\left ( 1-n \right )}\frac{dv}{dx}$

- wherein

$\frac{1}{y^{n}}\frac{dy}{dx}+\frac{p}{y^{n-1}}=Q$

Given equation can be written as

$\frac{dy}{dx} - 1/3 y = (x/6 ) y^4\Rightarrow 1/y^4 \frac{dy}{dx} - 1/3 y^3 = x/6$

Let $1/y^3 = t \Rightarrow -3/y^4 \frac{dy}{dx} = \frac{dt}{dx}$

$1/y^4 \frac{dy}{dx}= -1/3 \frac{dt }{dx}$

$(-1/3) \frac{dt}{dx} -t/3 = x/6 \Rightarrow \frac{dt}{dx}+ t = -x/2$

IF is $e^{\int 1dx }= e^x$

$\Rightarrow e^x \frac{dt}{dx} + e^x = \frac{-xe^x}{2} \Rightarrow \frac{d}{dx}(te^x) = -1/2 xe^x$

$\Rightarrow \int {d}(te^x) + 1/2\int xe^x=C$

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

$\Rightarrow 2+ (x-1)y^3 = 2cy^3e^{-x}$

Option 1)

$2+ (x+1)y^3 = 2cy^3e^x$

Option 2)

$2- (x+1)y^3 = 2cy^3e^{-x}`$

Option 3)

$2+ y^3(x-1) = 2cy^3e^{-x}$

Option 4)

$2- (x+1)y^3 = 2cy^3e^x$

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