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Solution of diffrential equation $\frac{dy}{dx} + y = y^{1/2}$ is

Option 1)
$\sqrt {xy}-1/3x^{3/2} = C$

Option 2)
$\sqrt {xy}+1/3x^{3/2} = C$

Option 3)
$\sqrt {xy}-1/2x^{3/2} = C$

Option 4)
$\sqrt {xy}+1/2x^{3/2} = C$

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

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

$dy/dx+ p(x) y = Qy^n$ is bernoulli's equation , on compairing this and given equation , we get p(x) = 1/x and Q(x)= 1

Now given equation can be written as

$\frac{1}{y^{1/2}}\frac{dy}{dx} + y^{1/2}\cdot 1/x$

Let ${y^{1/2}} = t \Rightarrow 1/ 2y^{1/2} \frac{dy}{dx}= \frac{dt}{dx}$

Equation becomes

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

It is of linear diffrential equation form , so IF $= e^{\int 1/2x}dx= e^{1/2lnx}= \sqrt x$

multiplying with $\sqrt x$ , both sides of equation

$\sqrt x \frac{dt}{dx} + (1/2\sqrt x )t= \sqrt x/ 2 \Rightarrow \frac{d}{dx}(\sqrt x \cdot t )= \frac{\sqrt x}{2 }$

$\Rightarrow \int {d}(\sqrt x \cdot t )-\int \frac{\sqrt x}{2 }dx = C \Rightarrow \sqrt xt - 1/3 x^{3/2} = c$

$\Rightarrow \sqrt{xy}-1/3x^{3/2}= C$

Bernoulli's Equation -

$\frac{dv}{dx}+\left (1-n \right )vP =Q\left ( 1-n \right )$

- wherein

This is the equation we get after solving bernoulli's equation.

Option 1)

$\sqrt {xy}-1/3x^{3/2} = C$

Option 2)

$\sqrt {xy}+1/3x^{3/2} = C$

Option 3)

$\sqrt {xy}-1/2x^{3/2} = C$

Option 4)

$\sqrt {xy}+1/2x^{3/2} = C$

Posted by

gaurav

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