Plus Two Maths Chapter 9 Differential Equations Chapter Wise Question and Answers PDF Download

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Plus Two Maths Chapter 9 Differential Equations Chapter Wise Question and Answers

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Board	Kerala Board
Study Materials	Chapter wise Question and Answers
For Year	2021
Class	12
Subject	Mathematics
Chapters	Maths Chapter 9 Differential Equations
Format	PDF
Provider	Spandanam Blog

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Question 1.
y = e^2x(a + bx), a and b are arbitrary constants.
Answer:
y = e^2x(a + bx) ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^2xb + (a + bx)2e^2x
\(\frac{d y}{d x}\) = 2y + be^2x ⇒ \(\frac{d y}{d x}\) – 2y = be^2x ____(2)
Differentiating (2) with respect to x,

Question 2.
y = e^x(acosx + 6sinx), a and b are arbitrary constants.
Answer:
y = e^x(acosx + 6sinx) ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^x(-asinx + bcosx) + e^x(acosx + bsinx) \(\frac{d y}{d x}\) = e^x(-asin x + b cos x) + y
\(\frac{d y}{d x}\) – y = e^x(-a sin x + b cos x) ____(2)
Differentiating (2) with respec to x,

Question 3.
y = c₁e^x + c₂ e^-x , c₁ and c₁ are arbitrary constants.
Answer:
y = c₁e^x + c₂ e^-x ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = c₁e^x + c₂ e^-x __(2)
Differentiating (2) with respect to x,

Question 4.
(x – a)² + 2y² = a², a is a arbitrary constants.
Answer:
(x – a)² + 2y² = a² ___(1)
Differentiating with respect to x,

Question 5.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Answer:
y\(\frac{d y}{d x}\) = x ⇒ ydy = xdx
Integrating on both sides,
∫ydy = ∫xdx + c
⇒ \(\frac{y^{2}}{2}=\frac{x^{2}}{2}\) + c ____(1)
Since it passes through (0, -2),

Question 6.
Form the DE representing the family of parabolas having a vertex at origin and axis along positive direction of x-axis.
Answer:
Let (a, 0) be focus of the given family of parabolas.
y² = 4ax ____(1)
Differentiating with respect to x,

Question 7.
For the DE xy \(\frac{d y}{d x}\) = (x + 2)(y + 2), find the solution curve passing through the point(1,- 1).
Answer:
xy \(\frac{d y}{d x}\) = (x + 2)(y + 2)
⇒ \(\frac{y}{y+2} d x=\frac{x+2}{x} d x\)
Integrating on both sides,

⇒ y – 2 log|y + 2| = x + 2log|x| + c ____(1)
Since it passes through (1, -1),
⇒ -1 – 2log|-1 + 2| = 1 + 2log|l| + c
⇒ -2 = c
(1) ⇒ y – 2log|y + 2| = x + 2log|x| – 2.

Question 8.
Solve the initial value problem: \(\frac{d y}{d x}\) = y tan 2x; y(0) = 2.
Answer:
\(\frac{d y}{d x}\) = y tan 2x
⇒ \(\frac{d y}{y}\) tan 2xdx,
This is a variable type
∴∫\(\frac{d y}{y}\) = ∫tan 2xdx ⇒ log y = \(\frac{1}{2}\) log|sec 2x| + c
Given y(0) = 2 ⇒ log 2 = \(\frac{1}{2}\) log|sec 0| + c ⇒ c = log 2
log y = \(\frac{1}{2}\) log|sec 2x| + log 2 ⇒

Plus Two Maths Differential Equations Four Mark Questions and Answers

Question 1.
(i) Consider the differential equation given below. (1)
\(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\). Write the order and degree of the DE (if defined)
(ii) Find the Differential equation satisfying the family of curves y² = a(b² – x²), a and b are arbitrary constants. (3)
Answer:
(i) 4; degree is not defined

(ii) y² = a(b² – x²) ____(1)
Differentiating with respect to x,
2y \(\frac{d y}{d x}\) = -a2x ⇒ y\(\frac{d y}{d x}\) = -ax ____(2)
Differentiating (2) with respect to x,

Which is the differential equation.

Question 2.

1. Find the Differential equation satisfying the family of curves y = ae^3x + be^-2x, a and b are arbitrary constants. (3)
2. Hence write the degree and order of the DE. (1)

Answer:
1. y = ae^3x + be^-2x ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = ae^3x × 3 + be^-2x × -2 ____(2)
Differentiating (2) with respect to x,
⇒ \(\frac{d^{2} y}{d x^{2}}\) = 9ae^3x + 4be^-2x ____(3)
Now, (3) + 2 × (2) ⇒ \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=15 a e^{3 x}\)

Using (4), (5) in (1), we have,

2. Order: 2; degree: 1.

Question 3.
Consider the equation of all circles which pass through the origin and whose centres are on the x-axis.

1. Define the general equation of the circle.(1)
2. Find the DE corresponding to the above equation. (3)

Answer:
1. The general equation of the circle, passing through the origin and whose centers lies on x-axis can be taken as (x – h)² + y² = h² where h being an arbitrary constant.

2. Simplifying (x – h)² + y² = h² we get,
x² – 2hx + h² + y² = h² ⇒ x² – 2hx + h² = 0 _____(1)
Differentiating we get,
2x + 2y \(\frac{d y}{d x}\) – 2h = 0 ⇒ h = x + y \(\frac{d y}{d x}\)
Substituting in (1) we can eliminate h

Question 4.
Find a particular solution satisfying the given condition. (x³ + x² + x +1)\(\frac{d y}{d x}\) = 2x² + x, y = 1, when x = 0.
Answer:

Integrating on both sides,
∫dy = ∫\(\frac{2 x^{2}+x}{\left(x^{2}+1\right)(x+1)} d x\)
Splitting into partial fractions we have, (see Unit:7)

Question 5.

1. Write the degree of the DE y’ = 2xy. [0, 1, 2, 3] (1)
2. Express y’ = 2xy in the form Mdx = Ndy. Where M is a function of x and N is the function of y. (2)
3. Solve y’ = 2xy, y(0) = 1 (1)

Answer:

1. Degree = 1

2. We have, \(\frac{d y}{d x}\) = 2xy ⇒ \(\frac{d y}{y}\) = 2xdx, which is of the form Mdx = Ndy.

3. Solution is ∫\(\frac{d y}{y}\) = 2∫xdx ⇒ log|y| = x² + c
Given y(0) = 1 ⇒ log|1| = 0 + c ⇒ c = 0
⇒ log|y| = x² ⇒ y = e^x2.

Question 6.
Solve the following DE \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\).
Answer:
\(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\), this is a Homogeneous DE.
Therefore, put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) to convert it into variable separable form.
The DE becomes,

Therefore integrating we get,

Question 7.
Solve the linear differential equation \(x \frac{d y}{d x}-y=(x-1) e^{x}\).
Answer:
Given, x\(\frac{d y}{d x}\) – y = (x – 1)e^x, dividing both sides by x ,we get

Solution is
y × IF = ∫Q(IF)dx + c

Question 8.
(i) Choose the correct answer from the bracket. The solution of the differential equation xdy + ydx = 0 represents (1)
(a) a rectangular hyperbola
(b) a parabola whose centre is origin
(c) a straight line whose centre is origin
(d) a circle whose centre is origin.
(ii) From the DE of the family of circles touching the x-axis at origin. (3)
Answer:
(i) (c) a straight line whose centre is origin.

(ii) Let (0, a) be the centre of the circle. Therefore the equation of the circle is
x² + (y – a)² = a²
⇒ x² + y² = 2ay
⇒ \(\frac{x^{2}+y^{2}}{y}\) = 2a ____(1)

Differentiating with respect to x,

Question 9.
Solve the DE x²\(\frac{d y}{d x}\) = x² – 2y² + xy.
Answer:
x²\(\frac{d y}{d x}\) = x² – 2y² + xy

this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Integrating on both sides,

Question 10.
Choose the correct answer from the bracket

1. The DE \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, x > 0 is of order …..[0,1,2,3] (1)
2. The integrating factor \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, is……..[x, e^x, -x, e^-x] (1)
3. Solve \(\frac{d y}{d x}+\frac{y}{x}\) = e^x (2)

Answer:
1. Order = 1

2. \(\frac{d y}{d x}+\frac{y}{x}\) = e^x is of the form \(\frac{d y}{d x}\) + Py = Q,
where P = \(\frac{1}{x}\), Q = e^x
IF = e^∫Pdx = e^{∫\(\frac{1}{x}\)dx} = e^logx = x

3. Solution is y.IF = ∫e^x. IFdx
⇒ yx = ∫x.e^xdx ⇒ yx = x.e^x – ∫e^xdx
⇒ yx = x.e^x – e^x + c ⇒ yx = e^x(x – 1) + c.

Question 11.
Solve the DE \(\frac{d y}{d x}=\frac{x+y}{x-y}\).
Answer:
\(\frac{d y}{d x}=\frac{x+y}{x-y}\), this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)

Integrating on both sides,

Plus Two Maths Differential Equations Six Mark Questions and Answers

Question 1.
Consider the DE \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

1. Identify the DE ? Give reason. (1)
2. Explain the method of solving the DE. (1)
3. Solve the DE. (4)

Answer:
1. Given DE is a Homogeneous DE. Since y³ + 3x²y and x³ + 3xy² are Homogeneous functions of same degree (deg = 3).

2. By giving a substitution y = v x and \(\frac{d y}{d x}\) = v + x\(\frac{d y}{d x}\)
we can convert the DE into variable separable.

3. Now we have, \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

∴ Integrating we get,
\(\int \frac{1+3 v^{2}}{2 v\left(1-v^{2}\right)} d v=\int \frac{d x}{x}\)

\(\frac{1}{2}\) log v – log(1 – v²) = log x + log c

Question 2.
Consider the D.E \(\frac{d y}{d x}+\frac{y}{x}=x^{2}\)

1. Find degree and order of DE . (1)
2. Solve the D.E. (4)
3. Find the particular solution when x = 1, y = 1. (1)

Answer:
1. Degree: 1, Order: 1.

2. The given D.E is first order linear DE of the form
\(\frac{d y}{d x}\) + Py = Q. Comparing we get, P = \(\frac{1}{x}\), Q = x²
∴ ∫Pdx = ∫\(\frac{1}{x}\)dx = logx
Integrating factor (I.F) = e^∫Pdx = e^logx = x
y.x = ∫x².xdx + c = ∫x³ dx + c
⇒ y.x = \(\frac{x^{4}}{4}\) + c ___(1)

3. Given, y = 1 when x = 1, then (1)
⇒ 1 × 1 = \(\frac{1}{4}\) + c ⇒ c = \(\frac{3}{4}\)
Therefore particular solution is
y.x = \(\frac{x^{4}}{4}\) + \(\frac{3}{4}\) ⇒ 4xy = x³ + 3.

Question 3.
Consider the equation.\(\frac{d y}{d x}\) + y = sin x

1. What is the order and degree of this equation? (1)
2. Find the integrating factor. (2)
3. Solve this equation. (3)

Answer:
1. Order = 1, Degree = 1

2. Given, \(\frac{d y}{d x}\) + y = sin x is of the form
\(\frac{d y}{d x}\) + Py = Q ⇒ P = 1, Q = sinx
Integrating factor = e^∫Pdx = e^∫1dx = e^x

3. Therefore solution is
y.IF = ∫Q.IFdx + c ⇒ ye^x = ∫e^x sinxdx + c ____(1)
∫sinx.e^xdx = e^x sinx – ∫cosx.e^xdx
= e^x sin x – cosx.e^x – ∫sinx.e^x dx
⇒ 2∫e^x sin xdx = e^x(sin x – cos x)
⇒ ∫e^x sinxdx = \(\frac{e^{x}}{2}\)(sinx – cosx)
(1) ⇒ ye^x = \(\frac{e^{x}}{2}\)(sinx – cosx) + c.

Question 4.
Considerthe D.E (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

1. Find \(\frac{d y}{d x}\), degree and order of the above differential equation. (1)
2. Find the integrating factor of the above differential equation. (2)
3. Solve the differential equation. (3)

Answer:
1. Given, (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

Here, Degree = 1, Order = 1.

2. The given DE is of the form \(\frac{d y}{d x}\) + Py = Q
Where,

Splitting it into partial fractions we get,

Put x = 1, ⇒ 6 = 2B ⇒ B = 3,
put x = -1, ⇒ 2 = -2A ⇒ A = -1

3. Solution is y × IF = ∫Q × IFdx + c

Question 5.
(i) The degree of the differential Equation \(\frac{d^{2} y}{d x^{2}}+\cos \left(\frac{d y}{d x}\right)=0\) is
(a) 2
(b) 1
(c) 0
(d) Not defined
(ii) Solve \(\frac{d y}{d x}\) + 2y tanx = sinx; y = 0, x = \(\frac{\pi}{3}\) (5)
Answer:
(i) (d) Not defined.

(ii) \(\frac{d y}{d x}\) + 2y tanx = sinx
Then, P = 2tanx, Q = sinx
IF = e^∫Pdx = e^∫2tanxdx = e^{2log sec x} = sec² x Solution is; y × IF = ∫Q(IF)dx + c
⇒ ysec² x = ∫sinx sec² xdx + c
⇒ ysec² x = ∫tanx secx dx + c
⇒ ysec²x = secx + c
Here; y = 0, x = \(\frac{\pi}{3}\)
⇒ 0 × sec² \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + c ⇒ c = -2
⇒ ysec² x = secx – 2.

Question 6.
(i) The order of the differential equation \(x^{4} \frac{d^{2} y}{d x^{2}}=1+\left(\frac{d y}{d x}\right)^{3}\) is
(a) 1
(b) 3
(c) 4
(d) 2
(ii) Find the particular solution of the (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1 (5)
Answer:
(i) (d) 2

(ii) (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1

⇒ 0(1 + 1²) = tan^-11 + c ⇒ c = \(-\frac{\pi}{4}\)
⇒ y(1 + x²) = an^-1x – \(\frac{\pi}{4}\).

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