Question

Let y_1 and y_2 be solution of a second order homogeneous linear differential equation y''+p(x)y'+q(x)=0, in R. Suppose that y_1(x)+y_2(x)=e^{-x}, W[y_1(x),y_2(x)]=e^x, where W[y_1,y_2] is the Wroian of y_1 and y_2. Find p(x), q(x) and the general form of y_1 and y_2.

Second order linear equations
ANSWERED
asked 2020-11-09
Let \(y_1\) and \(y_2\) be solution of a second order homogeneous linear differential equation \(y''+p(x)y'+q(x)=0\), in R. Suppose that \(y_1(x)+y_2(x)=e^{-x}\),
\(W[y_1(x),y_2(x)]=e^x\), where \(W[y_1,y_2]\) is the Wro
ian of \(y_1\) and \(y_2\).
Find p(x), q(x) and the general form of \(y_1\) and \(y_2\).

Answers (1)

2020-11-10
\(y''+p(x)y'+q(x)=0\) (*)
\(y_1(x)+y_2(x)=e^{-x}\)
\(W(y_1,y_2)=e^x\)
Since \(y_1\), and \(y_2\) are solution to (*)
So their linear combinations are also a solution to (*)
Hence \(y_1+y_2=e^{-x}\) is also a solution to (*)
Now,
\((y_1+y_2)'=-e^{-x}\)
\((y_1+y_2)""=e^{-x}\)
So, from (*) we have
\((y_1+y_2)''+p(x)(y_1+y_2)'+q(x)(y_1+y_2)=0\)
\(\Rightarrow e^{-x}-e^{-x}p(x)+e^{-x}q(x)=0\)
or \(e^{-x}[1-p(x)+q(x)]=0\)
As \(e^{-x}\neq0\ \forall\ x\)
Hence, \(1-p(x)+q(x)=0\) (I)
Now, \(W[y_1,y_2]=[(y_1,y_2),(y'_1,y''_2)]=y_1y'_2-y'_1y_2=e^x\) (II)
As \(y_1+y_2=e^{-x}\)
\(\Rightarrow y_2=e^{-x}-y_1\)
and \(y'_1+y'_2=-e^{-x}\)
so, \(y'_2=e^{-x}-y'_1\)
Putting the value of \(y_2\) and \(y'_2\) in (II) we got,
\(y_1(-e^{-x}-y'_1)-y'_1(e^{-x}-y_1)=e^x\)
\(\Rightarrow-e^{-x}y_1-y_1y'_1-e^{-x}y'_1+y_1y'_1=e^x\)
\(-e^{-x}[y_1+y'_1]=e^x\)
\(y_1+y'_1=-e^{2x}\)
\(\Rightarrow y'_1=-e^{2x}-y_1\)
\(\Rightarrow y''_1=-2e^{2x}-y_1\)
\(\Rightarrow y''_1=-2e^{2x}-y'_1=-2e^{2x}-(-e^{2x}-y_1)\)
\(=-e^{2x}+y_1\)
\(y''_1=-e^{2x}+y_1\)
Putting the value of \(y'_1\) and \(y''_1\) in (*) we get,
\((-e^{2x}+y_1)+p(x)(-e^{-2x}-y_1)+q(x)y_1=0\)
\(\Rightarrow-e^{2x}+y_1=p(x)y_1+q(x)y_1-e^{2x}p(x)=0\)
\(\Rightarrow-e^{2x}[1+p(x)]+y_1[1-p(x)+q(x)]=0\)
From (I) we have \(1-p(x)+q(x)=0\)
So,
\(-e^{2x}[1+p(x)]+y_1\cdot0=0\)
\(\Rightarrow e^{2x}[1+p(x)]=0\)
\(\Rightarrow1+p(x)=0\)
\(p(x)=-1\)
From (I) we have
\(1-(-1)+q(x)=0\)
\(\Rightarrow q(x)=-2\)
So, \(p(x)=-1\) and \(q(x)=-2\)
Hence the differential equation become,
\(y''-y'-2y=0\)
The characteristic equation is
\(v^2-v-2=0\)
or \(v^2-2v+v-2=0\)
\(v(v-2)+1(v-2)=0\)
so, \((v-2)(v+1)=0\)
\(\Rightarrow v=2\) or \(v=-1\)
Hence \(y_1(x)=e^{2x}\) and \(y_2=e^{-x}\)
Here \(y(x)=c_1e^{2x}+c_2e^{-x}\)
Answer: \(p(x)=-1,\ q(x)=-2,\ y(x)=c_1e^{2x}+c_2e^{-x}\)
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