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Home > CBSE > CBSE Important Questions Class 12 Maths Chapter 9 – Differential Equations

CBSE Important Questions Class 12 Maths Chapter 9 – Differential Equations

Class 12 Maths Chapter 9 Differential Equations is part of the CBSE Class 12 Maths syllabus. A differential equation is an equation that contains derivatives. Students learn the order and degree of a differential equation, how to form a differential equation from a given family of curves, and how to solve equations by the variable separable method, the homogeneous method and the linear (integrating factor) method.

These CBSE important questions for Class 12 Maths Chapter 9 cover finding the order and degree, the general and particular solution, solving by variable separation, solving homogeneous differential equations, the integrating factor for linear equations, and forming a differential equation from a family of curves. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define the order and the degree of a differential equation. Find the order and degree of (d²y/dx²)³ + (dy/dx)² + 5 = 0.

Answer: The order is the order of the highest derivative appearing in the equation.
The degree is the power of that highest order derivative, provided the equation is a polynomial in the derivatives.
For the given equation:
The highest derivative is d²y/dx², which is of order 2, so the order is 2.
Its power is 3, so the degree is 3.

Q2. What is the difference between the general solution and a particular solution?

Answer: The general solution of a differential equation contains as many arbitrary constants as the order of the equation. It represents a whole family of curves.
A particular solution is obtained from the general solution by giving specific values to the arbitrary constants, usually using given initial or boundary conditions. It represents one single curve from that family.
For example, the general solution of dy/dx = 2x is y = x² + C. If we are told that y = 5 when x = 1, then C = 4 and the particular solution is y = x² + 4.

Q3. Solve the differential equation dy/dx = (1 + y²)/(1 + x²).

Answer: This is a variable separable equation. Separate the variables:
dy/(1 + y²) = dx/(1 + x²)
Integrating both sides:
∫dy/(1 + y²) = ∫dx/(1 + x²)
tan⁻¹y = tan⁻¹x + C
So the general solution is
tan⁻¹y − tan⁻¹x = C

Q4. Solve dy/dx = e^{x + y}.

Answer: Write the right side as a product:
dy/dx = e^x · e^y
Separate the variables:
dy/e^y = e^x dx
e^−y dy = e^x dx
Integrating both sides:
−e^−y = e^x + C
Or rearranged:
e^x + e^−y = C (absorbing the sign into the constant)

Q5. What is a homogeneous differential equation and how is it solved?

Answer: A differential equation of the form dy/dx = f(x, y) is homogeneous if f(x, y) can be written as a function of y/x alone, that is f(λx, λy) = f(x, y).
Method of solving:
1. Put y = vx, so dy/dx = v + x (dv/dx).
2. Substitute into the equation. The variables v and x will now separate.
3. Integrate and then replace v by y/x.

Q6. Solve the homogeneous equation dy/dx = (x + y)/x.

Answer: Simplify first: dy/dx = 1 + y/x
Put y = vx, so dy/dx = v + x(dv/dx):
v + x (dv/dx) = 1 + v
x (dv/dx) = 1
dv = dx/x
Integrating:
v = log|x| + C
Replacing v by y/x:
y/x = log|x| + C
y = x(log|x| + C)

Q7. Write the standard form of a linear differential equation and the formula for the integrating factor.

Answer: The standard form of a linear differential equation of the first order is
dy/dx + Py = Q
where P and Q are functions of x alone (or constants).
The integrating factor is
IF = e^{∫P dx}
The solution is then
y × IF = ∫(Q × IF) dx + C
There is a similar form dx/dy + Px = Q when x is treated as the dependent variable.

Q8. Solve the differential equation dy/dx + 2y = e^−x.

Answer: This is linear with P = 2 and Q = e^−x.
Integrating factor: IF = e^{∫2 dx} = e^2x
Solution: y × e^2x = ∫e^−x × e^2x dx + C
= ∫e^x dx + C
= e^x + C
Dividing throughout by e^2x:
y = e^−x + C e^−2x

Q9. Form the differential equation of the family of curves y = mx, where m is an arbitrary constant.

Answer: There is one arbitrary constant, so the differential equation will be of order 1.
Differentiate y = mx with respect to x:
dy/dx = m
Now substitute m back into the original equation:
y = (dy/dx) x
So the differential equation is
x (dy/dx) − y = 0

Q10. Form the differential equation of the family of circles x² + y² = a², where a is an arbitrary constant.

Answer: Differentiate both sides with respect to x:
2x + 2y (dy/dx) = 0
Dividing by 2:
x + y (dy/dx) = 0
The constant a has disappeared on its own, so this is the required differential equation. It is of order 1, matching the single arbitrary constant.

Q11. Find the particular solution of dy/dx = y/x given that y = 2 when x = 1.

Answer: Separate the variables:
dy/y = dx/x
Integrating both sides:
log|y| = log|x| + log C
y = Cx
Applying the condition: y = 2 when x = 1 gives 2 = C(1), so C = 2.
The particular solution is y = 2x.

Q12. Solve x (dy/dx) + y = x³.

Answer: Divide throughout by x to get the standard linear form:
dy/dx + (1/x) y = x²
Here P = 1/x and Q = x².
Integrating factor: IF = e^{∫(1/x) dx} = e^{log x} = x
Solution: y × x = ∫x² × x dx + C
xy = ∫x³ dx + C
xy = x⁴/4 + C
Dividing by x:
y = x³/4 + C/x

Related Study Material on Extramarks

Q1.

$\begin{array}{l} Form a differential equation representing the family \\ of curves given by \frac{x}{a} + \frac{y}{b} = 1 . \end{array}$

Opt.

Ans.

$\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 \\ Differentiating both sides w . r . t . x, we get \\ \frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0 \\ Againg, differentiating both sides w . r . t . x, we get \\ \frac{1}{b} \frac{d^{2} y}{{dx}^{2}} = 0 \\ \frac{d^{2} y}{{dx}^{2}} = 0 y ‘ ‘ = 0 \\ Hence, the required differential equation is y = 0. \end{array}$

Q2.

$\begin{array}{l} Find the particular solution of the differential equation \\ \log (\frac{dy}{dx}) = 3 x + 4 y at (0) = 0 . \end{array}$

Opt.

$\begin{array}{l} \log (\frac{dy}{dx}) = 3 x + 4 y \frac{dy}{dx} = e^{3 x + 4 y} \\ \frac{dy}{dx} = e^{3 x} e^{4 y} \\ \frac{dy}{} \end{array}$

Ans.

$\begin{array}{l} \log (\frac{dy}{dx}) = 3 x + 4 y \frac{dy}{dx} = e^{3 x + 4 y} \\ \frac{dy}{dx} = e^{3 x} e^{4 y} \\ \frac{dy}{e^{4 y}} = e^{3 x} dx \\ Integrating both sides \\ « \frac{dy}{e^{4 y}} = « e^{3 x} dx \\ « e^{’ 4 y} dy = « e^{3 x} dx \\ \frac{e^{’ 4 y}}{’ 4} = \frac{e^{3 x}}{3} + c \\ Now at x = 0 and y = 0 \\ \frac{1}{’ 4} = \frac{1}{3} + c \\ ’ \frac{1}{3} ’ \frac{1}{4} = c \end{array}$

Q3.

$Solve: \frac{dy}{dx} = \frac{y}{x} e^{\frac{y}{x}}$

Opt.

$\begin{array}{l} \frac{dy}{dx} = \frac{y}{x} e^{\frac{y}{x}} \\ Let \frac{y}{x} = v \\ \frac{dy}{dx} = v + x \frac{dv}{dx} \end{array}$ $\begin{array}{l} \end{array}$

Ans.

$\begin{array}{l} \frac{dy}{dx} = \frac{y}{x} e^{\frac{y}{x}} \\ Let \frac{y}{x} = v \\ \frac{dy}{dx} = v + x \frac{dv}{dx} \\ v + x \frac{dv}{dx} = v e^{v} \\ x \frac{dv}{dx} = e^{v} \\ \frac{dv}{e} = \frac{dx}{x} \\ e^{’ v} dv = \frac{dx}{x} \\ Integrating both sides \\ « e^{v} dv = « \frac{dx}{x} \\ e^{’ v} = \log |x| + c \\ e^{\frac{’ y}{x}} = \log |x| + c \end{array}$

CBSE Important Questions Class 12 Maths Chapter 9 – Differential Equations

Related Study Material on Extramarks

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