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TN 12th Computer Applications மின்னணு தரவு பரிமாற்றம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிக பாதுகாப்பு அமைப்புகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

12th Standard English Medium Maths Subject Ordinary Differential Equations Creative 3 Mark Questions with Solution Part - I

Published on: 19/06/2021

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Std: 12th Standard - Maths

chapter: - Creative Questions

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Questions:

Generic 3

5 x 3 = 15

Find the D.E of all circles touching y-axis at the origin.

Form the D.E to y²=a(b-x)(b+x) by eliminating a and b as its parameters.

Form the D.E of the family of curves c(y + c)²= x², where c is the parameter.

Solve: x\(\frac{dy}{dx}\)+ 2y = x²

Form the differential equation for y = e^-2x [A cos 3x-B sin 3x]

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Answers Key:

Generic 3

5 x 3 = 15

\({ y }^{ 2 }-{ x }^{ 2 }=2xy\frac { dy }{ dx } \)

\(y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +\left( \frac { dy }{ dx } \right) ^{ 2 }=\frac { ydy }{ xdx } \)

\(8x\left( \frac { dy }{ dx } \right) ^{ 2 }-12y\left( \frac { dy }{ dx } \right) ^{ 2 }=27x\)

x\(\frac{dy}{dx}\)+2y = x²
\( \Rightarrow \frac { dy }{ dx } +\frac { 2y }{ x } =x\)
This is a linear differential equation
Here \(p=\frac { 2 }{ x } \)and Q = x
\(\int { pdx } =2\int { \frac { 1 }{ x } } =2logx={ logx }^{ 2 }\)
\(I.F={ e }^{ \int { pdx } }={ e }logx^{ 2 }={ x }^{ 2 }\)
∴ The solution is
\({ ye }^{ \int { dx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { x.{ x }^{ 2 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { { x }^{ 3 }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\frac { { x }^{ 2 } }{ 4 } +c\)

Given y = e-2x[A cos 3x- B sin 3x]
⇒ ye²x = A cos 3x-B sin 3x
Differentiating,y¹e²+2y e^2x= -3A sin 3x-3B
cos 3x
Differentiating again we get,
y"e^2x+2(2y')e^2x+4ye^2x= -9(A cos 3x-B sin 3x)
⇒ e^2x( y"+4y'+4y) = -9(A cos 3x - B sin 3x)
⇒ z y"+4y'+4y = -9(A cos 3x-B sin 3x)
⇒ y"+4y'+4y = -9(using (1))
⇒ y"+4y'+13y = 0
is the required differential equation