Mixing
Homogeneous or non - homogeneous $?$
$begingroup$
The second order differential equation is given by -
$ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$
Is this a homogeneous differential equation $?$
Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?
ordinary-differential-equations
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
The second order differential equation is given by -
$ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$
Is this a homogeneous differential equation $?$
Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?
ordinary-differential-equations
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
The second order differential equation is given by -
$ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$
Is this a homogeneous differential equation $?$
Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?
ordinary-differential-equations
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
$endgroup$
The second order differential equation is given by -
$ frac{d^{2}y}{dx^{2}} + sin (x+y) = sin x$
Is this a homogeneous differential equation $?$
Well, I guess this is not a homogeneous differential equation since the form of this equation is not $a(x)y'' + b(x)y' +c(x)y = 0$.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
asked Jan 24 at 18:33
MathsaddictMathsaddict
3669
3669
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $u=x+y$ ,
Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$
$dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$
$thereforedfrac{d^2u}{dx^2}+sin u=sin x$
This is a inhomogeneous differential equation.
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
add a comment |
$begingroup$
You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.
The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
Let $u=x+y$ ,
Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$
$dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$
$thereforedfrac{d^2u}{dx^2}+sin u=sin x$
This is a inhomogeneous differential equation.
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
add a comment |
$begingroup$
Let $u=x+y$ ,
Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$
$dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$
$thereforedfrac{d^2u}{dx^2}+sin u=sin x$
This is a inhomogeneous differential equation.
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
add a comment |
$begingroup$
Let $u=x+y$ ,
Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$
$dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$
$thereforedfrac{d^2u}{dx^2}+sin u=sin x$
This is a inhomogeneous differential equation.
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
$endgroup$
Let $u=x+y$ ,
Then $dfrac{du}{dx}=1+dfrac{dy}{dx}$
$dfrac{d^2u}{dx^2}=dfrac{d^2y}{dx^2}$
$thereforedfrac{d^2u}{dx^2}+sin u=sin x$
This is a inhomogeneous differential equation.
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
answered Jan 25 at 12:39
doraemonpauldoraemonpaul
12.7k31660
12.7k31660
add a comment |
add a comment |
$begingroup$
You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.
The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.
The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.
The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
$endgroup$
You are correct, as it is not a linear ODE, it is neither homogeneous nor inhomogeneous.
The cited characterization is most likely based on the fact that $y=0$ is a solution, but that is only a necessary condition for linearity, not a sufficient one.
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
answered Jan 25 at 13:00
LutzLLutzL
59.6k42057
59.6k42057
add a comment |
add a comment |
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