What is a solution to the differential equation $\frac{dy}{dy}=\sqrt{1-y}$?

rustenig
2022-09-14
Answered

What is a solution to the differential equation $\frac{dy}{dy}=\sqrt{1-y}$?

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asked 2022-05-26

I completed near all problems om a differential equations text chapter on reducing non-separable first order differential equations to separable by using an appropriate substitution for example u=y/x with y'=u+u'x and similar substations in y and x for making other similar problems separable.

I am not asking anyone to do the entire problem for me but I do need a little guidance to begin tackling this problem. I completely understand the procedure for making them separable. What I need help in is setting up the differential equation for the following word problem. Once this is done I can easily finish the problem.

Ch 1.4-29. Show that a straight line through the origin intersects all solution curves of a given differential equation y=g(y/x) at the same angle.

I am not asking anyone to do the entire problem for me but I do need a little guidance to begin tackling this problem. I completely understand the procedure for making them separable. What I need help in is setting up the differential equation for the following word problem. Once this is done I can easily finish the problem.

Ch 1.4-29. Show that a straight line through the origin intersects all solution curves of a given differential equation y=g(y/x) at the same angle.

asked 2022-06-29

I have been trying to solve a differential equation as a practice question for my test, but I am just unable to get the correct answer. Please have a look at the D.E:

$dy/dx=1/(3x+\mathrm{sin}(3y))$

My working is as follows:

$dx/dy=3x+\mathrm{sin}(3y)$

$dx-3x=\mathrm{sin}(3y)dy$

Integrating both sides:

$x-(3/2){x}^{2}=-(1/3)\mathrm{cos}(3y)+c$

But the correct answer is:

$x=c{e}^{3}y-1/6(\mathrm{cos}(3y)+\mathrm{sin}(3y))$, which is quite different from what I have got. Could someone please help me solve this?

$dy/dx=1/(3x+\mathrm{sin}(3y))$

My working is as follows:

$dx/dy=3x+\mathrm{sin}(3y)$

$dx-3x=\mathrm{sin}(3y)dy$

Integrating both sides:

$x-(3/2){x}^{2}=-(1/3)\mathrm{cos}(3y)+c$

But the correct answer is:

$x=c{e}^{3}y-1/6(\mathrm{cos}(3y)+\mathrm{sin}(3y))$, which is quite different from what I have got. Could someone please help me solve this?

asked 2022-06-22

I have a linear first order ordinary differential equation

$\frac{dy}{dx}+\mathrm{tan}(x)y=2{\mathrm{cos}}^{2}x\mathrm{sin}x-\mathrm{sec}x$

with an initial condition as $y(\frac{\pi}{4})=3\sqrt{2}$

My integrating factor $\mu (x)=\mathrm{sec}x$

After multiplication with the integration factor what I get is:

$(secx\text{}y{)}^{\prime}=sin2x-se{c}^{2}x$

or

$(secx\text{}y{)}^{\prime}=2sinxcosx-se{c}^{2}x$

If I use the first equation I get:

$y(x)=\frac{\frac{1}{2}cos2x-tanx+c}{secx}$

and using the second equation I get:

$y(x)=\frac{si{n}^{2}x-tanx+c}{secx}$

$\int 2\text{}sinx\text{}cosx\text{}dx=2\frac{si{n}^{2}x}{2}=si{n}^{2}x$

with the first equation I get c=7 and second equation I get $c=\frac{13}{2}$.

It is a very simple differential equation but when I solve it I get two different answers. Is this ok?

$\frac{dy}{dx}+\mathrm{tan}(x)y=2{\mathrm{cos}}^{2}x\mathrm{sin}x-\mathrm{sec}x$

with an initial condition as $y(\frac{\pi}{4})=3\sqrt{2}$

My integrating factor $\mu (x)=\mathrm{sec}x$

After multiplication with the integration factor what I get is:

$(secx\text{}y{)}^{\prime}=sin2x-se{c}^{2}x$

or

$(secx\text{}y{)}^{\prime}=2sinxcosx-se{c}^{2}x$

If I use the first equation I get:

$y(x)=\frac{\frac{1}{2}cos2x-tanx+c}{secx}$

and using the second equation I get:

$y(x)=\frac{si{n}^{2}x-tanx+c}{secx}$

$\int 2\text{}sinx\text{}cosx\text{}dx=2\frac{si{n}^{2}x}{2}=si{n}^{2}x$

with the first equation I get c=7 and second equation I get $c=\frac{13}{2}$.

It is a very simple differential equation but when I solve it I get two different answers. Is this ok?

asked 2022-04-10

Problem:

Solve the following differential equations.

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

Answer:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ \frac{dv}{dx}+3v& =3\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have P(x)=3.

$\begin{array}{rl}I(x)& ={e}^{3x}\\ {e}^{3x}\frac{dv}{dx}+3{e}^{3x}v& =3{e}^{3x}\\ D\left({e}^{3x}v\right)& =3{e}^{3x}\\ {e}^{3x}v& ={e}^{3x}+C\\ v& =1+{e}^{-3x}\\ {y}^{3}& =1+{e}^{-3x}\end{array}$

Now to check my answer.

$\begin{array}{rl}3{y}^{2}\frac{dy}{dx}& =-3C{e}^{-3x}\\ {y}^{2}\frac{dy}{dx}& =-C{e}^{-3x}\\ \frac{dy}{dx}& =-C{e}^{-3x}{y}^{-2}\\ x\frac{dy}{dx}+y& =x(-C{e}^{-3x}{y}^{-2})+{(1+{e}^{-3x})}^{\frac{1}{3}}\end{array}$

I cannot seem to get my answer to check. Where did I go wrong?

Here is my second attempt to solve the problem:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}$

Now, I want to write:

$D(3xv)=9$

but that is wrong. What did I do wrong?

Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& ={x}^{3}\\ {x}^{3}\frac{dv}{dx}+3{x}^{2}v& =3{x}^{2}\\ D({x}^{3}v)& ={x}^{3}+C\\ {x}^{3}v& ={x}^{3}+C\\ v& =C{x}^{-3}+1\\ {y}^{3}& =C{x}^{-3}+1\end{array}$

Do I have it right now?

Solve the following differential equations.

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

Answer:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ \frac{dv}{dx}+3v& =3\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have P(x)=3.

$\begin{array}{rl}I(x)& ={e}^{3x}\\ {e}^{3x}\frac{dv}{dx}+3{e}^{3x}v& =3{e}^{3x}\\ D\left({e}^{3x}v\right)& =3{e}^{3x}\\ {e}^{3x}v& ={e}^{3x}+C\\ v& =1+{e}^{-3x}\\ {y}^{3}& =1+{e}^{-3x}\end{array}$

Now to check my answer.

$\begin{array}{rl}3{y}^{2}\frac{dy}{dx}& =-3C{e}^{-3x}\\ {y}^{2}\frac{dy}{dx}& =-C{e}^{-3x}\\ \frac{dy}{dx}& =-C{e}^{-3x}{y}^{-2}\\ x\frac{dy}{dx}+y& =x(-C{e}^{-3x}{y}^{-2})+{(1+{e}^{-3x})}^{\frac{1}{3}}\end{array}$

I cannot seem to get my answer to check. Where did I go wrong?

Here is my second attempt to solve the problem:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}$

Now, I want to write:

$D(3xv)=9$

but that is wrong. What did I do wrong?

Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& ={x}^{3}\\ {x}^{3}\frac{dv}{dx}+3{x}^{2}v& =3{x}^{2}\\ D({x}^{3}v)& ={x}^{3}+C\\ {x}^{3}v& ={x}^{3}+C\\ v& =C{x}^{-3}+1\\ {y}^{3}& =C{x}^{-3}+1\end{array}$

Do I have it right now?

asked 2022-09-04

What is a general solution to the differential equation $y\prime =5{x}^{\frac{2}{3}}{y}^{4}$?

asked 2020-11-16

Solve the differential equation

asked 2022-07-02

I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab

${y}^{\prime}-xy=x{y}^{3/2}\phantom{\rule{thinmathspace}{0ex}},y(1)=4$

$\begin{array}{rlr}{y}^{\prime}-xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+{y}^{3/2})& \\ {\displaystyle \frac{dy}{(y+{y}^{3/2})}}=& x\phantom{\rule{thinmathspace}{0ex}}dx& \\ -2\mathrm{ln}{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{{x}^{2}}{2}}+c& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}+c}{)}^{2}}}& \\ 4=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-1}{4}}+c}{)}^{2}}}& \\ c=& -0.655& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}-0.655}{)}^{2}}}& \end{array}$

This is what I'm getting in Matlab

$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{{x}^{2}}{8}+\text{atanh}\left(3\right)-\frac{1}{8})+1)}^{2}}{4}\\ \frac{{(\text{tanh}(-\frac{{x}^{2}}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-1)}^{2}}{4}\end{array}\right)$

${y}^{\prime}-xy=x{y}^{3/2}\phantom{\rule{thinmathspace}{0ex}},y(1)=4$

$\begin{array}{rlr}{y}^{\prime}-xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+{y}^{3/2})& \\ {\displaystyle \frac{dy}{(y+{y}^{3/2})}}=& x\phantom{\rule{thinmathspace}{0ex}}dx& \\ -2\mathrm{ln}{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{{x}^{2}}{2}}+c& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}+c}{)}^{2}}}& \\ 4=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-1}{4}}+c}{)}^{2}}}& \\ c=& -0.655& \\ y=& {\displaystyle \frac{-1}{(1-{\mathrm{e}}^{{\textstyle \frac{-{x}^{2}}{4}}-0.655}{)}^{2}}}& \end{array}$

This is what I'm getting in Matlab

$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{{x}^{2}}{8}+\text{atanh}\left(3\right)-\frac{1}{8})+1)}^{2}}{4}\\ \frac{{(\text{tanh}(-\frac{{x}^{2}}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-1)}^{2}}{4}\end{array}\right)$