Solve $\frac{dy}{dx}=1+\frac{1}{{y}^{2}}$ ?

sombereki51
2022-09-26
Answered

Solve $\frac{dy}{dx}=1+\frac{1}{{y}^{2}}$ ?

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June Rowland

Answered 2022-09-27
Author has **6** answers

We have:

$\frac{dy}{dx}=1+\frac{1}{{y}^{2}}$

Which is a First Order Separable Ordinary Differential Equation so we can rearrange and "separate the variables":

$\frac{dy}{dx}=\frac{1+{y}^{2}}{{y}^{2}}$

$\Rightarrow \int \frac{{y}^{2}}{1+{y}^{2}}dy=\int dx$

We can manipulate the LHS integral:

$\int \frac{1+{y}^{2}-1}{1+{y}^{2}}dy=\int dx$

$\therefore \int 1-\frac{1}{1+{y}^{2}}dy=\int dx$

Which is now trivial to integrate giving us:

$y-\mathrm{arctan}\left(y\right)=x+C$

Which is the general implicit solution.

$\frac{dy}{dx}=1+\frac{1}{{y}^{2}}$

Which is a First Order Separable Ordinary Differential Equation so we can rearrange and "separate the variables":

$\frac{dy}{dx}=\frac{1+{y}^{2}}{{y}^{2}}$

$\Rightarrow \int \frac{{y}^{2}}{1+{y}^{2}}dy=\int dx$

We can manipulate the LHS integral:

$\int \frac{1+{y}^{2}-1}{1+{y}^{2}}dy=\int dx$

$\therefore \int 1-\frac{1}{1+{y}^{2}}dy=\int dx$

Which is now trivial to integrate giving us:

$y-\mathrm{arctan}\left(y\right)=x+C$

Which is the general implicit solution.

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asked 2022-05-15

How can I solve this ODE of first order:

$\begin{array}{rl}{y}^{{}^{\prime}}={y}^{2}+x,& \text{where}{y}^{{}^{\prime}}=\frac{dy}{dx}\end{array}$

Is there any exact mehod to solve it ?

Thanks!

$\begin{array}{rl}{y}^{{}^{\prime}}={y}^{2}+x,& \text{where}{y}^{{}^{\prime}}=\frac{dy}{dx}\end{array}$

Is there any exact mehod to solve it ?

Thanks!

asked 2022-06-25

I am given a single second order differential equation:

$\ddot{x}-{x}^{3}-2{x}^{2}\dot{x}+1=0$

.. and am asked to classify the critical points as stable, unstable or saddle points.

Finding the critical points is an easy task for first order differential equation(s), both single equation and system of equations. However, I have never done so for second order, and higher, equations.

I have an idea on how to solve it but not sure the approach is correct. Am I correct in saying I need to split the single differential equation into two differential equations and rename the relevant terms? Doing so on the above equation gives:

$\dot{{x}_{1}}={x}_{2}$

$\dot{{x}_{2}}-{x}_{1}^{3}-2{x}_{1}^{2}{x}_{2}+1=0$

Thereafter, I follow the same process as starting off with a set of two first order differential equations. That is, set ${\dot{x}}_{1}$ and ${\dot{x}}_{2}$ to 0, and solve for the intersection of ${x}_{1}$ and ${x}_{2}$ to find the fixed points. The nature of the fixed points would then be determined by calculating the Trace and Determinant of the Jacobian at the specific fixed points.

Is my thinking on the right track?

$\ddot{x}-{x}^{3}-2{x}^{2}\dot{x}+1=0$

.. and am asked to classify the critical points as stable, unstable or saddle points.

Finding the critical points is an easy task for first order differential equation(s), both single equation and system of equations. However, I have never done so for second order, and higher, equations.

I have an idea on how to solve it but not sure the approach is correct. Am I correct in saying I need to split the single differential equation into two differential equations and rename the relevant terms? Doing so on the above equation gives:

$\dot{{x}_{1}}={x}_{2}$

$\dot{{x}_{2}}-{x}_{1}^{3}-2{x}_{1}^{2}{x}_{2}+1=0$

Thereafter, I follow the same process as starting off with a set of two first order differential equations. That is, set ${\dot{x}}_{1}$ and ${\dot{x}}_{2}$ to 0, and solve for the intersection of ${x}_{1}$ and ${x}_{2}$ to find the fixed points. The nature of the fixed points would then be determined by calculating the Trace and Determinant of the Jacobian at the specific fixed points.

Is my thinking on the right track?

asked 2022-06-21

I have a first order differential equation question with initial value.

$\frac{dv}{dt}+\frac{{v}^{2}}{5}=9.8\text{with}v(0)=0$

I tried to solve it but just didn't know how to remove the absolute sign in my solution.

Please let me know if whole question is needed to remove the absolute sign inside the solution.

Edit 1: Part where I got stuck:

After integration and since C = 0:

$\frac{5ln(lv+7l)}{14}-\frac{5ln(lv-7l)}{14}=t$

Edit 2: I think my solution is wrong

$\frac{dv}{dt}+\frac{{v}^{2}}{5}=9.8\text{with}v(0)=0$

I tried to solve it but just didn't know how to remove the absolute sign in my solution.

Please let me know if whole question is needed to remove the absolute sign inside the solution.

Edit 1: Part where I got stuck:

After integration and since C = 0:

$\frac{5ln(lv+7l)}{14}-\frac{5ln(lv-7l)}{14}=t$

Edit 2: I think my solution is wrong

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Write the first order differential equation for