How to deal with the absolute value sign in the process of solving differential equation?
Imagine we are given a differential equation as follows
$$y'sin x-ycos x=0,$$
which seems to be very simple. We can solve it like this:
begin{align*}
&y'sin x-ycos x=0\
implies &frac{{rm d}y}{y}=cot x{rm d}x\
implies &int frac{{rm d}y}{y}=int cot x{rm d}x\
implies &ln y=lnsin x+C'\
implies &y=e^{lnsin x+C'}\
implies &y=Csin x.
end{align*}
Right? May be. But wait! Notice the transformation or operation from the the 3rd line to the 4th line. In general, we ought to write as these:
$$int frac{{rm d}y}{y}=ln |y|+C_1,~~~int cot x{rm d}x=ln|sin x|+C_2,$$
when we are finding the indefinite integral, namely, we should add an absolute value sign in the result.
Of course, sometimes, you can cancel the sign in the final result again, indeed. But some books boldly omit the sign in the process. This is irresponsible negligence, or a valid trick?
differential-equations
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
add a comment |
Imagine we are given a differential equation as follows
$$y'sin x-ycos x=0,$$
which seems to be very simple. We can solve it like this:
begin{align*}
&y'sin x-ycos x=0\
implies &frac{{rm d}y}{y}=cot x{rm d}x\
implies &int frac{{rm d}y}{y}=int cot x{rm d}x\
implies &ln y=lnsin x+C'\
implies &y=e^{lnsin x+C'}\
implies &y=Csin x.
end{align*}
Right? May be. But wait! Notice the transformation or operation from the the 3rd line to the 4th line. In general, we ought to write as these:
$$int frac{{rm d}y}{y}=ln |y|+C_1,~~~int cot x{rm d}x=ln|sin x|+C_2,$$
when we are finding the indefinite integral, namely, we should add an absolute value sign in the result.
Of course, sometimes, you can cancel the sign in the final result again, indeed. But some books boldly omit the sign in the process. This is irresponsible negligence, or a valid trick?
differential-equations
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
2
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58
add a comment |
Imagine we are given a differential equation as follows
$$y'sin x-ycos x=0,$$
which seems to be very simple. We can solve it like this:
begin{align*}
&y'sin x-ycos x=0\
implies &frac{{rm d}y}{y}=cot x{rm d}x\
implies &int frac{{rm d}y}{y}=int cot x{rm d}x\
implies &ln y=lnsin x+C'\
implies &y=e^{lnsin x+C'}\
implies &y=Csin x.
end{align*}
Right? May be. But wait! Notice the transformation or operation from the the 3rd line to the 4th line. In general, we ought to write as these:
$$int frac{{rm d}y}{y}=ln |y|+C_1,~~~int cot x{rm d}x=ln|sin x|+C_2,$$
when we are finding the indefinite integral, namely, we should add an absolute value sign in the result.
Of course, sometimes, you can cancel the sign in the final result again, indeed. But some books boldly omit the sign in the process. This is irresponsible negligence, or a valid trick?
differential-equations
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
Imagine we are given a differential equation as follows
$$y'sin x-ycos x=0,$$
which seems to be very simple. We can solve it like this:
begin{align*}
&y'sin x-ycos x=0\
implies &frac{{rm d}y}{y}=cot x{rm d}x\
implies &int frac{{rm d}y}{y}=int cot x{rm d}x\
implies &ln y=lnsin x+C'\
implies &y=e^{lnsin x+C'}\
implies &y=Csin x.
end{align*}
Right? May be. But wait! Notice the transformation or operation from the the 3rd line to the 4th line. In general, we ought to write as these:
$$int frac{{rm d}y}{y}=ln |y|+C_1,~~~int cot x{rm d}x=ln|sin x|+C_2,$$
when we are finding the indefinite integral, namely, we should add an absolute value sign in the result.
Of course, sometimes, you can cancel the sign in the final result again, indeed. But some books boldly omit the sign in the process. This is irresponsible negligence, or a valid trick?
differential-equations
differential-equations
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
edited Dec 27 '18 at 14:27
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
asked Dec 27 '18 at 8:46
mengdie1982
4,809618
4,809618
2
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58
add a comment |
2
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58
2
2
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58
add a comment |
1 Answer
1
active
oldest
votes
You deal with it like this
begin{align}
ln|y| &= ln|sin x| + C_2-C_1 \
|y| &= e^{C_2-C_1}|sin x| \
y &= pm e^{C_2-C_1}sin x
end{align}
Where one of the $pm$ signs is applied. You then rewrite $C = pm e^{C_2-C_1}$ to get the desired result. Clearly $C$ can be either positive or negative, depending on the given initial value.
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
add a comment |
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1 Answer
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You deal with it like this
begin{align}
ln|y| &= ln|sin x| + C_2-C_1 \
|y| &= e^{C_2-C_1}|sin x| \
y &= pm e^{C_2-C_1}sin x
end{align}
Where one of the $pm$ signs is applied. You then rewrite $C = pm e^{C_2-C_1}$ to get the desired result. Clearly $C$ can be either positive or negative, depending on the given initial value.
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
add a comment |
You deal with it like this
begin{align}
ln|y| &= ln|sin x| + C_2-C_1 \
|y| &= e^{C_2-C_1}|sin x| \
y &= pm e^{C_2-C_1}sin x
end{align}
Where one of the $pm$ signs is applied. You then rewrite $C = pm e^{C_2-C_1}$ to get the desired result. Clearly $C$ can be either positive or negative, depending on the given initial value.
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
add a comment |
You deal with it like this
begin{align}
ln|y| &= ln|sin x| + C_2-C_1 \
|y| &= e^{C_2-C_1}|sin x| \
y &= pm e^{C_2-C_1}sin x
end{align}
Where one of the $pm$ signs is applied. You then rewrite $C = pm e^{C_2-C_1}$ to get the desired result. Clearly $C$ can be either positive or negative, depending on the given initial value.
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
You deal with it like this
begin{align}
ln|y| &= ln|sin x| + C_2-C_1 \
|y| &= e^{C_2-C_1}|sin x| \
y &= pm e^{C_2-C_1}sin x
end{align}
Where one of the $pm$ signs is applied. You then rewrite $C = pm e^{C_2-C_1}$ to get the desired result. Clearly $C$ can be either positive or negative, depending on the given initial value.
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
answered Dec 27 '18 at 10:23
Dylan
12.3k31026
12.3k31026
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
add a comment |
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
$C$ could be $0$?
– mengdie1982
Dec 27 '18 at 14:28
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
Before anything happends, you should've pointed out that $y=0$ is a trivial solution. Then you go on to solve the equation for $yne 0$ (since you need to divide by $y$ in one step). In the end, the trivial solution happens to fit, so $C=0$ is included.
– Dylan
Dec 27 '18 at 16:10
add a comment |
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2
Absolute value signs are necessary. However, in certain intervals you can argue that not all choices of $pm$ signs are necessary. For example, if you want solutions in $(0,pi)$ there are two solutions $y=Csin, x$ and $y=-Csin, x$; $y$ cannot change sign within the interval.
– Kavi Rama Murthy
Dec 27 '18 at 8:55
See it as $C=sign(y_0),sign(sin(x_0))e^{C'}$, $y=0$ is a solution that can not be crossed by another solution, and the domain has to exclude the roots of the $sin x$, all signs are constant.
– LutzL
Dec 27 '18 at 8:58