What is the general solution of the differential equation ? $x\frac{dy}{dx}=\frac{2}{x}+2-y$ given that x=−1,y=0

Lina Neal
2022-09-11
Answered

What is the general solution of the differential equation ? $x\frac{dy}{dx}=\frac{2}{x}+2-y$ given that x=−1,y=0

You can still ask an expert for help

asked 2022-02-17

I am trying to solve a differential equation of the type $x{}^{\u2033}=-x+{x}^{3}$ . Now when I first integrate it with respect to time, $t$ , then I am getting ${x}^{\prime}=x({x}^{2}-1)+C$ , which is a non-linear first order differential equation. Now if the constant happens to be zero then I can solve it by partial fraction method but as life is not easier that constant term tagged to the $x}^{\prime$ is actually not zero. So how to proceed forward in this case.

Thank you!

Thank you!

asked 2020-11-16

Solve differential equation
$(6x+1){y}^{2}dy/dx+3{x}^{2}+2{y}^{3}=0$ , y(0)=1

asked 2022-09-12

What is a solution to the differential equation $y\frac{dy}{dx}={e}^{x}$ with y(0)=4?

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 2022-05-23

In my differential equations book, I have found the following:

Let ${P}_{0}(\frac{dy}{dx}{)}^{n}+{P}_{1}(\frac{dy}{dx}{)}^{n-1}+{P}_{2}(\frac{dy}{dx}{)}^{n-2}+......+{P}_{n-1}(\frac{dy}{dx})+{P}_{n}=0$ be the differential equation of first degree 1 and order n (where ${P}_{i}$ $\mathrm{\forall}$ i $\in 0,1,2,...n$ are functions of x and y).

Assuming that it is solvable for p, it can be represented as:

$[p-{f}_{1}(x,y)][p-{f}_{2}(x,y)][p-{f}_{3}(x,y)]........[p-{f}_{n}(x,y)]=0$

equating each factor to Zero, we get n differential equations of first order and first degree.

$[p-{f}_{1}(x,y)]=0,\text{}[p-{f}_{2}(x,y)]=0,\text{}[p-{f}_{3}(x,y)]=0,\text{}........[p-{f}_{n}(x,y)]=0$

Let the solution to these n factors be:

${F}_{1}(x,y,{c}_{1})=0,\text{}{F}_{2}(x,y,{c}_{2})=0,\text{}{F}_{3}(x,y,{c}_{3})=0,\text{}........{F}_{n}(x,y,{c}_{n})=0$

Where ${c}_{1},{c}_{2},{c}_{3}.....{c}_{n}$ are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace ${c}_{1},{c}_{2},{c}_{3}.....{c}_{n}$ by a single arbitrary constant c. Then the n solutions (4) can be re-written as

${F}_{1}(x,y,c)=0,\text{}{F}_{2}(x,y,c)=0,\text{}{F}_{3}(x,y,c)=0,\text{}........{F}_{n}(x,y,c)=0$

They can be combined to form the general solution as follows:

${F}_{1}(x,y,c)\text{}{F}_{2}(x,y,c)\text{}{F}_{3}(x,y,c)\text{}........{F}_{n}(x,y,c)=0\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}(1)$

Now, my question is, whether equation (1) is the most general form of solution to the differential equation.I think the following is the most general form of solution to the differential equation :

${F}_{1}(x,y,{c}_{1})\text{}{F}_{2}(x,y,{c}_{2})\text{}{F}_{3}(x,y,{c}_{3})\text{}........{F}_{n}(x,y,{c}_{n})=0\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}(2)$

If (1) is the general solution, the constant of integration can be found out by only one IVP say, $y(0)=0$. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.

Let ${P}_{0}(\frac{dy}{dx}{)}^{n}+{P}_{1}(\frac{dy}{dx}{)}^{n-1}+{P}_{2}(\frac{dy}{dx}{)}^{n-2}+......+{P}_{n-1}(\frac{dy}{dx})+{P}_{n}=0$ be the differential equation of first degree 1 and order n (where ${P}_{i}$ $\mathrm{\forall}$ i $\in 0,1,2,...n$ are functions of x and y).

Assuming that it is solvable for p, it can be represented as:

$[p-{f}_{1}(x,y)][p-{f}_{2}(x,y)][p-{f}_{3}(x,y)]........[p-{f}_{n}(x,y)]=0$

equating each factor to Zero, we get n differential equations of first order and first degree.

$[p-{f}_{1}(x,y)]=0,\text{}[p-{f}_{2}(x,y)]=0,\text{}[p-{f}_{3}(x,y)]=0,\text{}........[p-{f}_{n}(x,y)]=0$

Let the solution to these n factors be:

${F}_{1}(x,y,{c}_{1})=0,\text{}{F}_{2}(x,y,{c}_{2})=0,\text{}{F}_{3}(x,y,{c}_{3})=0,\text{}........{F}_{n}(x,y,{c}_{n})=0$

Where ${c}_{1},{c}_{2},{c}_{3}.....{c}_{n}$ are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace ${c}_{1},{c}_{2},{c}_{3}.....{c}_{n}$ by a single arbitrary constant c. Then the n solutions (4) can be re-written as

${F}_{1}(x,y,c)=0,\text{}{F}_{2}(x,y,c)=0,\text{}{F}_{3}(x,y,c)=0,\text{}........{F}_{n}(x,y,c)=0$

They can be combined to form the general solution as follows:

${F}_{1}(x,y,c)\text{}{F}_{2}(x,y,c)\text{}{F}_{3}(x,y,c)\text{}........{F}_{n}(x,y,c)=0\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}(1)$

Now, my question is, whether equation (1) is the most general form of solution to the differential equation.I think the following is the most general form of solution to the differential equation :

${F}_{1}(x,y,{c}_{1})\text{}{F}_{2}(x,y,{c}_{2})\text{}{F}_{3}(x,y,{c}_{3})\text{}........{F}_{n}(x,y,{c}_{n})=0\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}(2)$

If (1) is the general solution, the constant of integration can be found out by only one IVP say, $y(0)=0$. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.

asked 2022-09-14

What is a solution to the differential equation $\frac{dy}{dx}=4-6y$?

asked 2022-09-09

What is a solution to the differential equation $(y+1)\frac{dy}{dx}=2x$?