GATE 2023 AE – Question 54
For $x^2y\prime\prime+4xy\prime+2y=0$, x≥1, y(1)=0 and y′(1)=1, the value y(2) (two decimal places) is _____.
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Correct answer: 0.24 to 0.26
Explanation
The equation $x^2y^{\prime\prime}+4xy^\prime+2y=0$ is a Cauchy-Euler equation.
**Step 1: trial solution** $y=x^m$. Then $y^\prime=mx^{m-1}$ and $y^{\prime\prime}=m(m-1)x^{m-2}$:
$$m(m-1)+4m+2=0\;\Rightarrow\;m^2+3m+2=0\;\Rightarrow\;(m+1)(m+2)=0 .$$
So $m=-1$ or $m=-2$ and
$$y=\frac{C_1}{x}+\frac{C_2}{x^2}.$$
**Step 2: initial conditions.**
- $y(1)=C_1+C_2=0\;\Rightarrow\;C_2=-C_1$.
- $y^\prime=-\dfrac{C_1}{x^2}-\dfrac{2C_2}{x^3}$, so $y^\prime(1)=-C_1-2C_2=-C_1+2C_1=C_1=1$.
So $C_1=1$, $C_2=-1$ and $y=\dfrac1x-\dfrac1{x^2}$.
**Step 3:** $y(2)=\dfrac12-\dfrac14=\mathbf{0.25}$.