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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations3 Marks
Question
show that $\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}$ is a solution of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
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Answer

$\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}\ ...(1)$

Differentiating both sides with respect to x,

$​​​​\frac{\text{dy}}{\text{dx}}=\text{be}^\text{x}+2\text{ce}^{2\text{x}}\ ...(2)$

Differentiating both sides with respect to x,

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{be}^\text{x}+4\text{ce}^{2\text{x}}\ ...(3)$

now,

$\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}$

$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3(\text{be}^\text{x}+2\text{ce}^{2\text{x}})+2(\text{be}^\text{x}+\text{ce}^{2\text{x}})$

$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}+2\text{be}^\text{x}+2\text{ce}^{2\text{x}}$

$=3\text{be}^\text{x}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}-6\text{ce}^{2\text{x}}$

$=0$

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