Find the solutions of 4y'' = y satisfying (i) y→ 0 as x→+∞…
Find the solutions of $4y'' = y$ satisfying (i) $y\to 0$ as $x\to+\infty$ and $y'(0)=6$; (ii) $y\to 0$ as $x\to-\infty$ and $y'(0)=6$.
Explanation of Method
Selecting the decaying exponential branch under a slope condition at the origin
Step 1: The general solution
The characteristic equation is $4r^2-1=0$, so $r=\pm\tfrac12$, and the homogeneous equation has general solution
\[ y = C_1 e^{x/2} + C_2 e^{-x/2}, \qquad y' = \tfrac12 C_1 e^{x/2} - \tfrac12 C_2 e^{-x/2}. \]
Both boundary problems below select particular values of $C_1$ and $C_2$ from this two-parameter family — one by requiring decay at $+\infty$, the other by requiring decay at $-\infty$.
Step 2: Decay as $x\to+\infty$, with $y'(0)=6$
Condition 1°.
As $x\to+\infty$, the term $e^{x/2}$ blows up while $e^{-x/2}\to 0$. For $y\to 0$ we therefore need $C_1=0$, leaving $y=C_2e^{-x/2}$ and $y'=-\tfrac12C_2e^{-x/2}$. The condition $y'(0)=6$ gives
\[ -\tfrac12 C_2 = 6 \quad\Longrightarrow\quad C_2=-12. \]
\[y = -12\,e^{-x/2}\]
Check: $y'=6e^{-x/2}$, so $y'(0)=6$; as $x\to+\infty$, $e^{-x/2}\to0$ so $y\to0$; and $4y''=4\cdot(-12)\cdot\tfrac14 e^{-x/2}=-12e^{-x/2}=y$, so the differential equation is satisfied as well.
Step 3: Decay as $x\to-\infty$, with $y'(0)=6$
Condition 2°.
This time it is $e^{-x/2}$ that diverges as $x\to-\infty$, while $e^{x/2}\to0$. Decay forces $C_2=0$, leaving $y=C_1e^{x/2}$ and $y'=\tfrac12C_1e^{x/2}$. Then $y'(0)=6$ gives
\[ \tfrac12 C_1 = 6 \quad\Longrightarrow\quad C_1=12. \]
\[y = 12\,e^{x/2}\]
Check: $y'=6e^{x/2}$, so $y'(0)=6$; as $x\to-\infty$, $e^{x/2}\to0$ so $y\to0$; and $4y''=4\cdot12\cdot\tfrac14e^{x/2}=12e^{x/2}=y$, confirming the equation holds.
Result
Final answer:
\[\boxed{y = -12\,e^{-x/2}}\]
Final answer:
\[\boxed{y = 12\,e^{x/2}}\]
Remark
Remark. The two conditions are mirror images of each other under $x\mapsto -x$, and so are the two solutions: replacing $x$ by $-x$ in $y=-12e^{-x/2}$ gives $-12e^{x/2}$, which is the negative of the second solution — the sign difference comes entirely from $y'(0)=6$ being an odd condition under that reflection.