Hugh Soto

2022-09-12

What is a particular solution to the differential equation $\frac{dy}{dx}=\frac{4\sqrt{y}\mathrm{ln}x}{x}$ with y(e)=1?

Clarence Mills

Beginner2022-09-13Added 18 answers

We have:

$\frac{dy}{dx}=\frac{4\sqrt{y}\mathrm{ln}x}{x}$

Which is a first order linear separable Differential Equation, so we can rearrange to get:

$\frac{1}{\sqrt{y}}\frac{dy}{dx}=\frac{4\mathrm{ln}x}{x}$

and separate the variables to get:

$\int {y}^{-\frac{1}{2}}dy=\int \frac{4\mathrm{ln}x}{x}dx$

And then we can integrate to get:

$\frac{{y}^{\frac{1}{2}}}{\frac{1}{2}}=\left(4\right)\left(\frac{{\mathrm{ln}}^{2}x}{2}\right)+C$

$\therefore 2\sqrt{y}=2{\mathrm{ln}}^{2}x+C$

Using y(e)=1 we get:

$\therefore 2=2{\mathrm{ln}}^{2}e+C$

$\therefore C=0$

Hence the particular solution is:

$2\sqrt{y}=2{\mathrm{ln}}^{2}x$

$\therefore \sqrt{y}={\mathrm{ln}}^{2}x$

$\therefore y={\mathrm{ln}}^{4}x$

Validation:

1. $x=e\Rightarrow y={\mathrm{ln}}^{4}e=1$ QED

2. $\frac{dy}{dx}=\frac{4{\mathrm{ln}}^{3}x}{x}=\frac{4\mathrm{ln}x}{x}\cdot {\mathrm{ln}}^{2}x=\frac{4\sqrt{y}\mathrm{ln}x}{x}$ QED

$\frac{dy}{dx}=\frac{4\sqrt{y}\mathrm{ln}x}{x}$

Which is a first order linear separable Differential Equation, so we can rearrange to get:

$\frac{1}{\sqrt{y}}\frac{dy}{dx}=\frac{4\mathrm{ln}x}{x}$

and separate the variables to get:

$\int {y}^{-\frac{1}{2}}dy=\int \frac{4\mathrm{ln}x}{x}dx$

And then we can integrate to get:

$\frac{{y}^{\frac{1}{2}}}{\frac{1}{2}}=\left(4\right)\left(\frac{{\mathrm{ln}}^{2}x}{2}\right)+C$

$\therefore 2\sqrt{y}=2{\mathrm{ln}}^{2}x+C$

Using y(e)=1 we get:

$\therefore 2=2{\mathrm{ln}}^{2}e+C$

$\therefore C=0$

Hence the particular solution is:

$2\sqrt{y}=2{\mathrm{ln}}^{2}x$

$\therefore \sqrt{y}={\mathrm{ln}}^{2}x$

$\therefore y={\mathrm{ln}}^{4}x$

Validation:

1. $x=e\Rightarrow y={\mathrm{ln}}^{4}e=1$ QED

2. $\frac{dy}{dx}=\frac{4{\mathrm{ln}}^{3}x}{x}=\frac{4\mathrm{ln}x}{x}\cdot {\mathrm{ln}}^{2}x=\frac{4\sqrt{y}\mathrm{ln}x}{x}$ QED

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