- #1
zenterix
- 675
- 83
- Homework Statement
- Consider the differential equation
$$y''+y'^2=0$$
- Relevant Equations
- a) Find all solutions.
b) Verify that ##1## and ##\log{x}## are linearly independent solutions on any interval to the right of the origin. Is ##c_1+c_2\log{x}## the general solution. If not, why not?
a) We can use reduction of order
$$p=y'\tag{1}$$
$$p'=y''\tag{2}$$
The DE becomes
$$p'+p^2=0\tag{3}$$
$$\frac{1}{p^2}p'=-1\tag{4}$$
This last step contains the assumption that ##p^2=y'^2\neq 0##.
$$-\left (\frac{1}{p(x)}-\frac{1}{p(x_i)}\right )=-(x-x_i)\tag{5}$$
$$p(x)=\frac{p(x_i)}{1+p(x_i)(x-x_i)}=y'(x)\tag{6}$$
$$y'(x)=\frac{y'(x_i)}{1+y'(x_i)(x-x_i)}\tag{7}$$
$$=\frac{1}{\frac{1}{y'(x_i)}+x-x_i}\tag{8}$$
$$y(x)=\ln{\left ( \frac{1}{y'(x_i)}+x-x_i \right )} + C\tag{9}$$
b) We can easily verify that ##1## and ##\log{x}## are solutions and that they are linearly independent (their ratio is not constant).
My question is about why ##c_1+c_2\log{x}## is not the general solution.
A few initial observations and mini-questions.
First of all, when we say "general solution" are we implicitly saying "general solution on interval ##[a,b]##"?
That is, can there be a general solution on ##[a_1,b_1]## and different general solution on ##[a_2,b_2]##?
From (9) we can see that there are solutions that are defined for certain intervals with ##x<0##.
Here is one argument I came up with to show that ##c_1+c_2\log{x}## is not a general solution.
Suppose we have two solutions ##y_1## and ##y_2##.
Is ##y_1+y_2## a solution?
$$(y_1''+y_2'')+(y_1'+y_2')^2=(y_1''+y_1'^2)+(y_2''+y_2'^2)+2y_1'y_2'\tag{10}$$
$$=2y_1'y_2'\tag{11}$$
Thus, ##y_1+y_2## is a solution only if ##y_1'y_2'=0##.
This is not always the case. For example, ##y_1(x)=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}## are both solutions but ##y_1'y_2'=\frac{1}{x(1+x)}\neq 0##.
If ##c_1+c_2\ln{x}## were a general solution then we'd have
$$y_1=a_1+a_2\ln{x}\tag{12}$$
$$y_2=b_1+b_2\ln{x}\tag{13}$$
and
$$y_1+y_2=(a_1+b_1)+(a_2+b_2)\ln{x}\tag{14}$$
would be another solution. But our example shows that this isn't true for ##y_1=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}##.
$$p=y'\tag{1}$$
$$p'=y''\tag{2}$$
The DE becomes
$$p'+p^2=0\tag{3}$$
$$\frac{1}{p^2}p'=-1\tag{4}$$
This last step contains the assumption that ##p^2=y'^2\neq 0##.
$$-\left (\frac{1}{p(x)}-\frac{1}{p(x_i)}\right )=-(x-x_i)\tag{5}$$
$$p(x)=\frac{p(x_i)}{1+p(x_i)(x-x_i)}=y'(x)\tag{6}$$
$$y'(x)=\frac{y'(x_i)}{1+y'(x_i)(x-x_i)}\tag{7}$$
$$=\frac{1}{\frac{1}{y'(x_i)}+x-x_i}\tag{8}$$
$$y(x)=\ln{\left ( \frac{1}{y'(x_i)}+x-x_i \right )} + C\tag{9}$$
b) We can easily verify that ##1## and ##\log{x}## are solutions and that they are linearly independent (their ratio is not constant).
My question is about why ##c_1+c_2\log{x}## is not the general solution.
A few initial observations and mini-questions.
First of all, when we say "general solution" are we implicitly saying "general solution on interval ##[a,b]##"?
That is, can there be a general solution on ##[a_1,b_1]## and different general solution on ##[a_2,b_2]##?
From (9) we can see that there are solutions that are defined for certain intervals with ##x<0##.
Here is one argument I came up with to show that ##c_1+c_2\log{x}## is not a general solution.
Suppose we have two solutions ##y_1## and ##y_2##.
Is ##y_1+y_2## a solution?
$$(y_1''+y_2'')+(y_1'+y_2')^2=(y_1''+y_1'^2)+(y_2''+y_2'^2)+2y_1'y_2'\tag{10}$$
$$=2y_1'y_2'\tag{11}$$
Thus, ##y_1+y_2## is a solution only if ##y_1'y_2'=0##.
This is not always the case. For example, ##y_1(x)=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}## are both solutions but ##y_1'y_2'=\frac{1}{x(1+x)}\neq 0##.
If ##c_1+c_2\ln{x}## were a general solution then we'd have
$$y_1=a_1+a_2\ln{x}\tag{12}$$
$$y_2=b_1+b_2\ln{x}\tag{13}$$
and
$$y_1+y_2=(a_1+b_1)+(a_2+b_2)\ln{x}\tag{14}$$
would be another solution. But our example shows that this isn't true for ##y_1=\ln{(1+x)}## and ##y_2(x)=\ln{(e(1+x))}##.
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