Khaleesi Herbert
2020-11-08
Answered

Solve differential equation$x{y}^{\prime}=(1-{y}^{2}{)}^{\frac{1}{2}}$

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asked 2020-11-14

Solve and classify equation
$dy/dx=-({e}^{x}+y)/(k+x+y{e}^{x})$ , y(0)=1

asked 2022-07-08

I'm writing down the differential equation that I was solving, but it should be relatable to any other first-order linear differential equation. The equation is:

$x\frac{dy}{dx}={x}^{2}+3y$ where $x>0$.

At first I got the wrong integrating factor. I realized it could be wrong after verification of the solution found and after I had made sure that I did not make any algebraic errors. Then I changed the integrating factor and got the correct solution to the equation. However, I'm still not sure why the first integrating factor does not work, and why is the second method is the way to go to find the correct solution.

Here is what I did:

(1) Wrong Integrating Factor:

$v(x)={e}^{\int -3{x}^{-1}dx}={e}^{-3\int {x}^{-1}dx}={e}^{-3\mathrm{ln}x}={e}^{-3}{e}^{\mathrm{ln}x}=\frac{x}{{e}^{3}}$

With this integrating factor we find the solution to be $y=\frac{{x}^{2}}{3}+C{x}^{-1}$ which is wrong.

(2) Correct Integrating Factor:

$v(x)={e}^{\int -3{x}^{-1}dx}={e}^{-3\int {x}^{-1}dx}={e}^{-3\mathrm{ln}x}={e}^{\mathrm{ln}{x}^{-3}}={x}^{-3}$

With this integrating factor we find the correct solution $y=-{x}^{2}+C{x}^{3}$

$x\frac{dy}{dx}={x}^{2}+3y$ where $x>0$.

At first I got the wrong integrating factor. I realized it could be wrong after verification of the solution found and after I had made sure that I did not make any algebraic errors. Then I changed the integrating factor and got the correct solution to the equation. However, I'm still not sure why the first integrating factor does not work, and why is the second method is the way to go to find the correct solution.

Here is what I did:

(1) Wrong Integrating Factor:

$v(x)={e}^{\int -3{x}^{-1}dx}={e}^{-3\int {x}^{-1}dx}={e}^{-3\mathrm{ln}x}={e}^{-3}{e}^{\mathrm{ln}x}=\frac{x}{{e}^{3}}$

With this integrating factor we find the solution to be $y=\frac{{x}^{2}}{3}+C{x}^{-1}$ which is wrong.

(2) Correct Integrating Factor:

$v(x)={e}^{\int -3{x}^{-1}dx}={e}^{-3\int {x}^{-1}dx}={e}^{-3\mathrm{ln}x}={e}^{\mathrm{ln}{x}^{-3}}={x}^{-3}$

With this integrating factor we find the correct solution $y=-{x}^{2}+C{x}^{3}$

asked 2022-06-22

An ODE (Ordinary Differential Equation) of order n becomes a relation:

$F(x,y,{y}^{(1)},...,{y}^{(n)})=0$

Then $F(x,y,{y}^{(1)})=0$ defines an ODE of order one. In "basic standard texts", for purposes of simplicity, is assumed that some ODE of first order can take the form:

${y}^{(1)}=f(x,y)$

for certain suitable f. Here is my "silly" question: What if that assumpion is not possible?

For example how I can deal with equations of the form:

${({y}^{(1)})}^{5}+sen({y}^{(1)})+{e}^{{y}^{(1)}}+x=0$

I appreciate any reference. Thanks in advance for your comments!

$F(x,y,{y}^{(1)},...,{y}^{(n)})=0$

Then $F(x,y,{y}^{(1)})=0$ defines an ODE of order one. In "basic standard texts", for purposes of simplicity, is assumed that some ODE of first order can take the form:

${y}^{(1)}=f(x,y)$

for certain suitable f. Here is my "silly" question: What if that assumpion is not possible?

For example how I can deal with equations of the form:

${({y}^{(1)})}^{5}+sen({y}^{(1)})+{e}^{{y}^{(1)}}+x=0$

I appreciate any reference. Thanks in advance for your comments!

asked 2022-06-14

${x}^{\prime}=\frac{t-x}{t},t>0$

I tried to solve it by:

${x}^{\prime}=1-\frac{x}{t}$

$\int dx=\int (1-\frac{x}{t})dt$

$x=t-x\mathrm{ln}t+C$

$x=\frac{t}{1+lnt}+C$

But this doesn't seem right. What am I doing wrong? Also I'm supposed to determine the maximal interval around t=1, but how do I get rid of $C$?

I tried to solve it by:

${x}^{\prime}=1-\frac{x}{t}$

$\int dx=\int (1-\frac{x}{t})dt$

$x=t-x\mathrm{ln}t+C$

$x=\frac{t}{1+lnt}+C$

But this doesn't seem right. What am I doing wrong? Also I'm supposed to determine the maximal interval around t=1, but how do I get rid of $C$?

asked 2021-12-26

What is the derivative of $h\left(x\right)=\mathrm{sin}2x\mathrm{cos}2x$

asked 2021-01-13

Solve differential equation

asked 2022-09-30

What is the solution of the differential equation? : $y\frac{dy}{dx}={(9-4{y}^{2})}^{\frac{1}{2}}$ where x=0 when y=0