# 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.

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 Wro
ian of $$y_1$$ and $$y_2$$.
Find p(x), q(x) and the general form of $$y_1$$ and $$y_2$$.

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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