How to find xdydx=y+4x?It's given that x&gt;1,y=1 when x=1.

You want to solve xy′=y+4x, which is equivalent toxy′−y=4xxy′−yx2=4x3Can you notice derivative of some familiar expression there? Are you able to continue from there?

Integrating factors I can explain. You may be familiar with the product rule for derivatives.d(uv)=udv+uduThe idea behind an integrating factor is to get one side of the equation to look like (uy)′=uy′+u′y, at which point you'll be able to integrate both sides. First, let's get all y's on one side.xy′−y=4xy′−yx=4x2Now we haveu′u=−1xln⁡u=−ln⁡x=ln⁡1xu=1xIf you multiply both sides by 1x, you'll get what Martin has in his answer. You can verify that the left side is equal to (yx)′.

As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem xz′=zx first which has z=x as a solution and then doing variation of parameters on y:=v(x)z(x), which gives:x(v′x+v)=vx+4xwhich should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part, xy′−y. That's why variation of parameters will always work.

You want to solve xy′=y+4x, which is equivalent toxy′−y=4xxy′−yx2=4x3Can you notice derivative of some familiar expression there? Are you able to continue from there?

Integrating factors I can explain. You may be familiar with the product rule for derivatives.d(uv)=udv+uduThe idea behind an integrating factor is to get one side of the equation to look like (uy)′=uy′+u′y, at which point you'll be able to integrate both sides. First, let's get all y's on one side.xy′−y=4xy′−yx=4x2Now we haveu′u=−1xln⁡u=−ln⁡x=ln⁡1xu=1xIf you multiply both sides by 1x, you'll get what Martin has in his answer. You can verify that the left side is equal to (yx)′.

As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem xz′=zx first which has z=x as a solution and then doing variation of parameters on y:=v(x)z(x), which gives:x(v′x+v)=vx+4xwhich should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part, xy′−y. That's why variation of parameters will always work.

"How to find xdydx=y+4x?It's given that x>1,y=1 when" solved and explained

Pam Stokes

Answered question

2022-01-20

How to find

x \frac{d y}{d x} = y + \frac{4}{x}

?
It's given that

x > 1, y = 1 when x = 1

Answer & Explanation

macalpinee3

Beginner2022-01-20Added 29 answers

You want to solve

x y^{'} = y + \frac{4}{x}

, which is equivalent to

x y^{'} - y = \frac{4}{x}

\frac{x y^{'} - y}{x^{2}} = \frac{4}{x^{3}}

Can you notice derivative of some familiar expression there? Are you able to continue from there?

godsrvnt0706

Beginner2022-01-21Added 31 answers

Integrating factors I can explain. You may be familiar with the product rule for derivatives.

d (u v) = u d v + u d u

The idea behind an integrating factor is to get one side of the equation to look like

{(u y)}^{'} = u y^{'} + u^{'} y

, at which point you'll be able to integrate both sides. First, let's get all y's on one side.

x y^{'} - y = \frac{4}{x}

y^{'} - \frac{y}{x} = \frac{4}{x^{2}}

Now we have

\frac{u^{'}}{u} = - \frac{1}{x}

\ln u = - \ln x = \ln \frac{1}{x}

u = \frac{1}{x}

If you multiply both sides by

\frac{1}{x}

, you'll get what Martin has in his answer. You can verify that the left side is equal to

{(\frac{y}{x})}^{'}

RizerMix

Expert2022-01-27Added 656 answers

As an alternative to the other answer, it is always worthwile to try and solve the homogenous problem

x z^{'} = \frac{z}{x}

first which has

z = x

as a solution and then doing variation of parameters on

y := v (x) z (x)

, which gives:

x (v^{'} x + v) = v x + \frac{4}{x}

which should simplify nicely. Note that this takes advantage of the fact that your equation consists of a homogeneous part,

x y^{'} - y

. That's why variation of parameters will always work.

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