# GATE Papers >> Mechanical >> 2018 >> Question No 146

Question No. 146

Given the ordinary differential equation

$\frac{d^2y}{dx^2}+\frac{dy}{dx}-6y=0$

with $y\left(0\right)=0$ and $\frac{dy}{dx}\left(0\right)=1$ , the value of $y\left(1\right)$ is _________ (correct to two decimal places).

##### Answer : 1.45 to 1.48

Solution of Question No 146 of GATE 2018 Mechanical Paper

$\frac{\mathrm d^2\mathrm y}{\mathrm{dx}^2}+\frac{\mathrm{dy}}{\mathrm{dx}}-6\mathrm y=0$

The auxiliary equation is : $\left(\mathrm D^2+\mathrm D-6\right)y=0$

$\begin{array}{l}\left(\mathrm D+3\right)\left(\mathrm D-2\right)=0\\\therefore\mathrm D=-3,2\end{array}$

The solution is $\mathrm y={\mathrm C}_1\mathrm e^{-3\mathrm x}+{\mathrm C}_2\mathrm e^{2\mathrm x}$

$\therefore\frac{\mathrm{dy}}{\mathrm{dx}}=-3{\mathrm C}_1\mathrm e^{-3\mathrm x}+2{\mathrm C}_2\mathrm e^{2\mathrm x}$

Given that,

When $\mathrm x=0,\;\mathrm y=0$

$0={\mathrm C}_1+{\mathrm C}_2.............(\mathrm i)$

Also, when $\mathrm x=0\;;\;\frac{\mathrm{dy}}{\mathrm{dx}}=1$

$-3{\mathrm C}_1+{\mathrm C}_2=1...............(\mathrm{ii})$

From (i) & (ii)

$\begin{array}{l}{\mathrm C}_2=\frac15,\;{\mathrm C}_1=-\frac15\\\mathrm y=\frac15\mathrm e^{2\mathrm x}-\frac15\mathrm e^{-3\mathrm x}\\\mathrm{When}\;\mathrm x=1\\\mathrm y\left(1\right)=\frac15\mathrm e^2-\frac15\mathrm e^{-3}\\\mathrm y\left(1\right)=\frac15\left(\mathrm e^2-\mathrm e^{-3}\right)=\frac{7.34}5=1.468\end{array}$

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