How do you solve for the equation $\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2}y}$ that satisfies the initial condition f(0)=12?

sooxicyiy
2022-09-14
Answered

How do you solve for the equation $\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2}y}$ that satisfies the initial condition f(0)=12?

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Hofpoetb9

Answered 2022-09-15
Author has **17** answers

First of all, I think ther is a mistake in your writing, I think you wanted to write:

$\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2y}}$

This is a separable differential equations, so:

${e}^{2y}dy=3{x}^{2}dx\Rightarrow \int {e}^{2y}dy=\int 3{x}^{2}dx\Rightarrow$

$\frac{1}{2}{e}^{2y}={x}^{3}+c$.

Now to find c let's use the condition: $f\left(0\right)=\frac{1}{2}$

$\frac{1}{2}{e}^{2\cdot \frac{1}{2}}={0}^{3}+c\Rightarrow c=\frac{1}{2}e$

So the solution is:

$\frac{1}{2}{e}^{2y}={x}^{3}+\frac{1}{2}e\Rightarrow {e}^{2y}=2{x}^{3}+e\Rightarrow 2y=\mathrm{ln}(2{x}^{3}+e)\Rightarrow$

$y=\frac{1}{2}\mathrm{ln}(2{x}^{3}+e)$

$\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2y}}$

This is a separable differential equations, so:

${e}^{2y}dy=3{x}^{2}dx\Rightarrow \int {e}^{2y}dy=\int 3{x}^{2}dx\Rightarrow$

$\frac{1}{2}{e}^{2y}={x}^{3}+c$.

Now to find c let's use the condition: $f\left(0\right)=\frac{1}{2}$

$\frac{1}{2}{e}^{2\cdot \frac{1}{2}}={0}^{3}+c\Rightarrow c=\frac{1}{2}e$

So the solution is:

$\frac{1}{2}{e}^{2y}={x}^{3}+\frac{1}{2}e\Rightarrow {e}^{2y}=2{x}^{3}+e\Rightarrow 2y=\mathrm{ln}(2{x}^{3}+e)\Rightarrow$

$y=\frac{1}{2}\mathrm{ln}(2{x}^{3}+e)$

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but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{{x}^{2}+1}-x$.

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$\begin{array}{}\text{(1)}& A\frac{\mathrm{\partial}}{\mathrm{\partial}t}g(t)+Bg(t)=f(t),\end{array}$

where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that $A\to 0$

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$\begin{array}{}\text{(1)}& A\frac{\mathrm{\partial}}{\mathrm{\partial}t}g(t)+Bg(t)=f(t),\end{array}$

where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that $A\to 0$

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I know there is an exact solution to Eq. (1), which is

$g(t)=C{e}^{-Bt/A}+\frac{1}{A}{\int}_{-\mathrm{\infty}}^{t}{e}^{-B(t-{t}^{\prime})/A}f({t}^{\prime})d{t}^{\prime},$

where C=0 because g(t)=0 if f(t)=0. However, I do not understand how this exact solution reduces to the case where A=0, which is $g(t)={B}^{-1}f(t)$. Any insight would be greatly appreciated.

I've seen a lot of documents discussing asymptotic analyses of linear differential equations, but they all start with second-order equations. Is this because there is inherently problematic with first-order?

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