# 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.

Let ${y}_{1}$ and ${y}_{2}$ be solution of a second order homogeneous linear differential equation ${y}^{″}+p\left(x\right){y}^{\prime }+q\left(x\right)=0$, in R. Suppose that ${y}_{1}\left(x\right)+{y}_{2}\left(x\right)={e}^{-x}$,
$W\left[{y}_{1}\left(x\right),{y}_{2}\left(x\right)\right]={e}^{x}$, where $W\left[{y}_{1},{y}_{2}\right]$ is the Wro
ian of ${y}_{1}$ and ${y}_{2}$.
Find p(x), q(x) and the general form of ${y}_{1}$ and ${y}_{2}$.
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delilnaT
${y}^{″}+p\left(x\right){y}^{\prime }+q\left(x\right)=0$ (*)
${y}_{1}\left(x\right)+{y}_{2}\left(x\right)={e}^{-x}$
$W\left({y}_{1},{y}_{2}\right)={e}^{x}$
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,
$\left({y}_{1}+{y}_{2}{\right)}^{\prime }=-{e}^{-x}$
$\left({y}_{1}+{y}_{2}\right)""={e}^{-x}$
So, from (*) we have
$\left({y}_{1}+{y}_{2}{\right)}^{″}+p\left(x\right)\left({y}_{1}+{y}_{2}{\right)}^{\prime }+q\left(x\right)\left({y}_{1}+{y}_{2}\right)=0$
$⇒{e}^{-x}-{e}^{-x}p\left(x\right)+{e}^{-x}q\left(x\right)=0$
or ${e}^{-x}\left[1-p\left(x\right)+q\left(x\right)\right]=0$
As
Hence, $1-p\left(x\right)+q\left(x\right)=0$ (I)
Now, $W\left[{y}_{1},{y}_{2}\right]=\left[\left({y}_{1},{y}_{2}\right),\left({y}_{1}^{\prime },{y}_{2}^{″}\right)\right]={y}_{1}{y}_{2}^{\prime }-{y}_{1}^{\prime }{y}_{2}={e}^{x}$ (II)
As ${y}_{1}+{y}_{2}={e}^{-x}$
$⇒{y}_{2}={e}^{-x}-{y}_{1}$
and ${y}_{1}^{\prime }+{y}_{2}^{\prime }=-{e}^{-x}$
so, ${y}_{2}^{\prime }={e}^{-x}-{y}_{1}^{\prime }$
Putting the value of ${y}_{2}$ and ${y}_{2}^{\prime }$ in (II) we got,
${y}_{1}\left(-{e}^{-x}-{y}_{1}^{\prime }\right)-{y}_{1}^{\prime }\left({e}^{-x}-{y}_{1}\right)={e}^{x}$
$⇒-{e}^{-x}{y}_{1}-{y}_{1}{y}_{1}^{\prime }-{e}^{-x}{y}_{1}^{\prime }+{y}_{1}{y}_{1}^{\prime }={e}^{x}$
$-{e}^{-x}\left[{y}_{1}+{y}_{1}^{\prime }\right]={e}^{x}$
${y}_{1}+{y}_{1}^{\prime }=-{e}^{2x}$
$⇒{y}_{1}^{\prime }=-{e}^{2x}-{y}_{1}$
$⇒{y}_{1}^{″}=-2{e}^{2x}-{y}_{1}$
$⇒{y}_{1}^{″}=-2{e}^{2x}-{y}_{1}^{\prime }=-2{e}^{2x}-\left(-{e}^{2x}-{y}_{1}\right)$
$=-{e}^{2x}+{y}_{1}$
${y}_{1}^{″}=-{e}^{2x}+{y}_{1}$
Putting the value of ${y}_{1}^{\prime }$ and ${y}_{1}^{″}$ in (*) we get,
$\left(-{e}^{2x}+{y}_{1}\right)+p\left(x\right)\left(-{e}^{-2x}-{y}_{1}\right)+q\left(x\right){y}_{1}=0$