\(y''+p(x)y'+q(x)=0\) (*)

\(y_1(x)+y_2(x)=e^{-x}\)

\(W(y_1,y_2)=e^x\)

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: \(p(x)=-1,\ q(x)=-2,\ y(x)=c_1e^{2x}+c_2e^{-x}\)

\(y_1(x)+y_2(x)=e^{-x}\)

\(W(y_1,y_2)=e^x\)

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: \(p(x)=-1,\ q(x)=-2,\ y(x)=c_1e^{2x}+c_2e^{-x}\)