CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 07

CBSE Class 12 Mathematics Linear Differential Equations (7). The Relations And Functions questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practice them to clear their Relations And Functions concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Relations And Functions worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Relations And Functions chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the D.E. of all the parabolas with latus rectum \( 4a \) and whose axes are parallel to \( x \)-axis.
Answer: Equation of parabola is given by \[ (y - k)^2 = 4a(x - h) \dots\dots (i) \] Where \( h, k \) are parameters Diff. w.r.t \( x \)

\(\Rightarrow 2(y - k)\frac{dy}{dx} = 4a(1 - 0) \)
\(\Rightarrow (y - k)\frac{dy}{dx} = 2a \dots\dots (ii) \) Diff. w.r.t. \( x \)
\(\Rightarrow (y - k)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \dots\dots (iii) \) From eq. (ii) \( (y - k) = \frac{2a}{\frac{dy}{dx}} \) put in eq. (iii)
\(\Rightarrow \left(\frac{2a}{\frac{dy}{dx}}\right) \left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 = 0 \)
\(\Rightarrow 2a\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 = 0 \) ans.

Question. Show that the D.E. representing one parameter family of curves \( (x^2 - y^2) = c (x^2 + y^2)^2 \) is \( (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \).
Answer: We have, \( (x^2 - y^2) = c(x^2 + y^2)^2 \dots\dots (i) \) Diff. w.r.t. \( x \),
\(\Rightarrow 2x - 2y\frac{dy}{dx} = 2c(x^2 + y^2) \cdot \left(2x + 2y\frac{dy}{dx}\right) \)
\(\Rightarrow x - y\frac{dy}{dx} = 2c(x^2 + y^2) \cdot \left(x + y\frac{dy}{dx}\right) \) Put value of \( c = \frac{x^2 - y^2}{(x^2 + y^2)^2} \) from (i) in equation (ii)
\(\Rightarrow x - y\frac{dy}{dx} = 2\frac{(x^2 - y^2)}{(x^2 + y^2)^2} \cdot (x^2 + y^2) \left(x + y\frac{dy}{dx}\right) \)
\(\Rightarrow (x^2 + y^2) \left(x - y\frac{dy}{dx}\right) = 2(x^2 - y^2) \left(x + y\frac{dy}{dx}\right) \)
\(\Rightarrow x^3 - x^2y\frac{dy}{dx} + y^2x - y^3\frac{dy}{dx} = 2x^3 + 2x^2y\frac{dy}{dx} - 2y^2x - 2y^3\frac{dy}{dx} \)
\(\Rightarrow \frac{dy}{dx} \left(-x^2y - y^3 - 2x^2y + 2y^3\right) = 2x^3 - 2y^2x - x^3 - y^2x \)
\(\Rightarrow \frac{dy}{dx} \left(y^3 - 3x^2y\right) = x^3 - 3y^2x \)
\(\Rightarrow (y^3 - 3x^2y)dy = (x^3 - 3y^2x)dx \) (proved)

Question. Find the D.E. of all non-vertical lines in a plane.
Answer: Let equation of line is given by
\(\Rightarrow \frac{x}{a} + \frac{y}{b} = 1 \dots\dots (i) \) where \( a \) & \( b \) are parameters Diff. w.r.t. \( x \)
\(\Rightarrow \frac{1}{a} + \frac{1}{b}\frac{dy}{dx} = 0 \dots\dots (ii) \) Diff. again w.r.t. \( x \)
\(\Rightarrow 0 + \frac{1}{b}\frac{d^2y}{dx^2} = 0 \)
\(\Rightarrow \frac{d^2y}{dx^2} = 0 \) ans.

Question. Show that \( xy = ae^x + be^{-x} + x^2 \) is a solution of the D.E. \( x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2 = 0 \).
Answer: We have, \( xy = ae^x + be^{-x} + x^2 \dots\dots (i) \) Diff. w.r.t. \( x \)
\(\Rightarrow x\frac{dy}{dx} + y = ae^x - be^{-x} + 2x \) Diff. again w.r.t \( x \)
\(\Rightarrow x\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) = ae^x + be^{-x} + 2 \)
\(\Rightarrow x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = xy - x^2 + 2 \quad \{ \text{from eq. (i) } ae^x + be^{-x} = xy - x^2 \} \)
\(\Rightarrow x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2 = 0 \)
\( \therefore \) the given function is a solution of the given D.E. ans.

Question. Verify that the function \( y = c_1e^{ax} \cos(bx) + c_2e^{ax} \sin(bx) \); \( c_1 \) & \( c_2 \) are arbitrary constants is a, solution of the D.E. \( \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \).
Answer: We have, \( y = c_1e^{ax} \cos(bx) + c_2e^{ax} \sin(bx) \)
\(\Rightarrow y = e^{ax} \cdot (c_1 \cos(bx) + c_2 \sin(bx)) \dots\dots (i) \) Diff. w.r.t. \( x \)
\(\Rightarrow \frac{dy}{dx} = e^{ax}(-bc_1 \sin(bx) + bc_2 \cos(bx)) + (c_1 \cos(bx) + c_2 \sin(bx)) \cdot e^{ax} \cdot a \)
\(\Rightarrow \frac{dy}{dx} = e^{ax} \cdot (-bc_1 \sin(bx) + bc_2 \cos(bx)) + ay \dots\dots \{ \text{from eq. (i)} \} \dots\dots (ii) \) Diff. again w.r.t. \( x \)
\(\Rightarrow \frac{d^2y}{dx^2} = e^{ax} (-b^2c_1 \cos(bx) - b^2c_2 \sin(bx)) + (-bc_1 \sin(bx) + bc_2 \cos(bx)) \cdot e^{ax} \cdot a + a\frac{dy}{dx} \)
\(\Rightarrow \frac{d^2y}{dx^2} = -b^2e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)) + \left(\frac{dy}{dx} - ay\right) a + a\frac{dy}{dx} \dots\dots \{ \text{from eq. (ii)} \} \)
\(\Rightarrow \frac{d^2y}{dx^2} = -b^2y + a\frac{dy}{dx} - a^2y + a\frac{dy}{dx} \)
\(\Rightarrow \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \)
Hence the given function is the solution of the given differential equation. ans.

Question. Show that \( y = cx + \frac{a}{c} \) is a solution of the D.E. \( y = x\frac{dy}{dx} + \frac{a}{\frac{dy}{dx}} \).
Answer: We have, \( y = cx + \frac{a}{c} \dots\dots (i) \) Diff. \( \frac{dy}{dx} = c \) Taking RHS
\( x\frac{dy}{dx} + \frac{a}{\frac{dy}{dx}} \)
\( = xc + \frac{a}{c} \)
\( = y \dots\dots (\text{from eq. (i)}) = \text{LHS} \)
\( \therefore \) given function is a solution of the given D.E.

Question. Show that the function defined by \( y = \sin x - \cos x, x \in R \) is a solution of the initial value problem \( \frac{dy}{dx} = \sin x + \cos x ; y(0) = -1 \).
Answer: We have, \( y = \sin x - \cos x \) Diff. w.r.t. \( x \)
\( \frac{dy}{dx} = \cos x + \sin x \), which is the given D.E.
Thus, \( y = \sin x - \cos x \) satisfies the D.E., hence it is a solution.
Also, when \( x = 0 \); \( y = \sin 0 - \cos 0 = -1 \) i.e., \( y(0) = -1 \)
Hence, \( y = \sin x - \cos x \) is a solution of the given initial value problem.

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