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The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
A differential equation links a function to its own rate of change, and solving one means recovering the family of functions that satisfies that link. This chapter classifies equations by order and degree, shows how a relation with arbitrary constants gives rise to a differential equation, and develops two systematic solution methods — variable separable and the integrating-factor technique for linear equations — that cover the vast majority of exam problems.
The order and degree of the differential equation (d²y/dx²)² + (dy/dx)³ = 0 are:
Answer
Order 2, degree 2. The highest derivative present is d²y/dx², a second-order derivative, so the order is 2. The degree is the power to which this highest-order derivative is raised, once the equation is polynomial in derivatives; here (d²y/dx²)² shows it is raised to the power 2, so the degree is 2.
The integrating factor of the differential equation dy/dx + y = eˣ is:
Answer
The integrating factor is eˣ. Here P = 1 (the coefficient of y), so IF = e^∫P dx = e^∫1 dx = eˣ.
The general solution of dy/dx = eˣ is:
Answer
y = eˣ + C. Integrating both sides directly: y = ∫eˣ dx = eˣ + C, where C is an arbitrary constant.
The differential equation obtained by eliminating the arbitrary constant c from y = cx² is:
Answer
The differential equation is x dy/dx = 2y. Differentiating y = cx² gives dy/dx = 2cx. From the original relation, c = y/x². Substituting: dy/dx = 2x(y/x²) = 2y/x, which rearranges to x dy/dx = 2y.
Assertion (A): The differential equation dy/dx = (x+y)/x is homogeneous. Reason (R): A differential equation dy/dx = f(x,y) is homogeneous if f(x,y) can be written entirely as a function of y/x.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the standard criterion for homogeneity correctly. For Assertion A, f(x,y) = (x+y)/x = 1 + y/x, which is indeed a function of y/x alone, so the equation satisfies exactly the criterion given in R, confirming both A and the explanatory role of R.
Find the order and degree of the differential equation d³y/dx³ + 2(d²y/dx²)² + dy/dx = 0.
Answer
The highest-order derivative present is d³y/dx³, a third-order derivative, so the order is 3. Since d³y/dx³ appears to the first power (it is not squared or raised to any other power), the degree is 1.
Solve the differential equation dy/dx = y/x.
Answer
This is variable separable. Rewrite as dy/y = dx/x. Integrating both sides: ∫dy/y = ∫dx/x gives ln|y| = ln|x| + C1. Exponentiating: y = e^(C1)·x = kx, where k is an arbitrary constant. So the general solution is y = kx.
Form the differential equation representing the family of curves y = Ae^(2x), where A is an arbitrary constant.
Answer
Differentiate y = Ae^(2x) with respect to x: dy/dx = 2Ae^(2x). Since Ae^(2x) = y (from the original equation), substitute: dy/dx = 2y. This is the required differential equation, with the single arbitrary constant A eliminated.
Solve the linear differential equation dy/dx + y/x = x², for x > 0.
Answer
This is a first-order linear equation with P = 1/x and Q = x². Find the integrating factor: IF = e^∫(1/x)dx = e^(ln x) = x. Multiply through by the integrating factor and use the general solution formula: y·x = ∫x²·x dx = ∫x³ dx = x⁴/4 + C. So the general solution is y·x = x⁴/4 + C, which can be written as y = x³/4 + C/x.
Solve the differential equation x(dy/dx) = x + y, given the initial condition y(1) = 1.
Answer
Rewrite as dy/dx = 1 + y/x, which is homogeneous since the right side depends only on y/x. Substitute y = vx, so dy/dx = v + x(dv/dx). The equation becomes v + x(dv/dx) = 1 + v, which simplifies to x(dv/dx) = 1. Separate variables: dv = dx/x. Integrating: v = ln|x| + C. Substitute back v = y/x: y/x = ln|x| + C, so y = x(ln|x| + C). Apply the initial condition y(1) = 1: 1 = 1×(ln 1 + C) = 1×(0 + C), so C = 1. The particular solution is y = x(ln|x| + 1) = x ln|x| + x.
Read the following and answer the questions that follow: A population P(t) grows at a rate proportional to the population present, modelled by the differential equation dP/dt = kP, where k is a positive constant. (a) Solve the differential equation to express P(t) in terms of the initial population P(0) and k. (b) If the population doubles every 10 years, find k in terms of ln 2. (c) If the initial population is 1000, find the population after 10 years. (d) Find the population after 20 years.
Answer
(a) Separate variables: dP/P = k dt. Integrating: ln P = kt + C, so P(t) = P0·e^(kt), where P0 = P(0). (b) Doubling in 10 years means P0·e^(10k) = 2P0, so e^(10k) = 2, giving 10k = ln 2, hence k = (ln 2)/10. (c) With P0 = 1000: P(10) = 1000·e^(10k) = 1000·e^(ln2) = 1000×2 = 2000. (d) P(20) = 1000·e^(20k) = 1000·e^(2 ln2) = 1000×2² = 1000×4 = 4000.
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