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What is thedifferential equation whose solution represents the family y = ae^3x + be^x?(a) d^2y/dx^2 – 3dy/dx + 4y = 0(b) d^2y/dx^2 – 4dy/dx + 3y = 0(c) d^2y/dx^2 + 4dy/dx + 3y = 0(d) d^2y/dx^2 – 4dy/dx – 3y = 0The question was posed to me in a national level competition.I want to ask this question from Linear Second Order Differential Equations topic in section Differential Equations of Mathematics – Class 12

Answer»

The correct option is (b) d^2y/dx^2 – 4dy/dx + 3y = 0

For explanation: The original given equation is,

y = ae^3x + be^x ……….(1)

DIFFERENTIATING the above equation, we get

dy/dx = 3ae^3x + be^x ……….(2)

d^2y/dx^2 = 9ae^3x + be^x ……….(3)

Now, to obtain the final equation we have to make the RHS = 0,

So, to get RHS = 0, we MULTIPLY, (1) with 3 and (2) with -4,

Thus , 3y = 3ae^3x + 3be^x ……….(4)

And -4(dy/dx) = -12ae^3x – 4 be^x……….(5)

Now, adding the above three EQUATIONS,(i.e. (3) + (4) + (5)) we get,

d^2y/dx^2 – 4dy/dx + 3y = 0


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