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## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 9 Differential Equations

### Plus Two Maths Differential Equations Three Mark Questions and Answers

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^{2x}b + (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 = c_{1}e^{x} + c_{2} e^{-x} , c_{1} and c_{1} are arbitrary constants.

Answer:

y = c_{1}e^{x} + c_{2} e^{-x} ___(1)

Differentiating with respect to x,

\(\frac{d y}{d x}\) = c_{1}e^{x} + c_{2} e^{-x} __(2)

Differentiating (2) with respect to x,

Question 4.

(x – a)^{2} + 2y^{2} = a^{2}, a is a arbitrary constants.

Answer:

(x – a)^{2} + 2y^{2} = a^{2} ___(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.

y^{2} = 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 y^{2} = a(b^{2} – x^{2}), a and b are arbitrary constants. (3)

Answer:

(i) 4; degree is not defined

(ii) y^{2} = a(b^{2} – x^{2}) ____(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.

- Find the Differential equation satisfying the family of curves y = ae
^{3x}+ be^{-2x}, a and b are arbitrary constants. (3) - 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.

- Define the general equation of the circle.(1)
- 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} + y^{2} = h^{2} where h being an arbitrary constant.

2. Simplifying (x – h)^{2} + y^{2} = h^{2} we get,

x^{2} – 2hx + h^{2} + y^{2} = h^{2} ⇒ x^{2} – 2hx + h^{2} = 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. (x^{3} + x^{2} + x +1)\(\frac{d y}{d x}\) = 2x^{2} + 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.

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

Given y(0) = 1 ⇒ log|1| = 0 + c ⇒ c = 0

⇒ log|y| = x^{2} ⇒ 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

x^{2} + (y – a)^{2} = a^{2}

⇒ x^{2} + y^{2} = 2ay

⇒ \(\frac{x^{2}+y^{2}}{y}\) = 2a ____(1)

Differentiating with respect to x,

Question 9.

Solve the DE x^{2}\(\frac{d y}{d x}\) = x^{2} – 2y^{2} + xy.

Answer:

x^{2}\(\frac{d y}{d x}\) = x^{2} – 2y^{2} + 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

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

⇒ 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 \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

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

Answer:

1. Given DE is a Homogeneous DE. Since y^{3} + 3x^{2}y and x^{3} + 3xy^{2} 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 – v^{2}) = log x + log c

Question 2.

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

- Find degree and order of DE . (1)
- Solve the D.E. (4)
- 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 = x^{2}

∴ ∫Pdx = ∫\(\frac{1}{x}\)dx = logx

Integrating factor (I.F) = e^{∫Pdx} = e^{logx} = x

y.x = ∫x^{2}.xdx + c = ∫x^{3} 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 = x^{3} + 3.

Question 3.

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

- What is the order and degree of this equation? (1)
- Find the integrating factor. (2)
- 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^{x}dx = e^{x} sinx – ∫cosx.e^{x}dx

= 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 (x^{2} – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

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

Answer:

1. Given, (x^{2} – 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} = sec^{2} x Solution is; y × IF = ∫Q(IF)dx + c

⇒ ysec^{2} x = ∫sinx sec^{2} xdx + c

⇒ ysec^{2} x = ∫tanx secx dx + c

⇒ ysec^{2}x = secx + c

Here; y = 0, x = \(\frac{\pi}{3}\)

⇒ 0 × sec^{2} \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + c ⇒ c = -2

⇒ ysec^{2} 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 + x^{2}) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1 (5)

Answer:

(i) (d) 2

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

⇒ 0(1 + 1^{2}) = tan^{-1}1 + c ⇒ c = \(-\frac{\pi}{4}\)

⇒ y(1 + x^{2}) = an^{-1}x – \(\frac{\pi}{4}\).