Class 12 Mathematics Chapter 9 Differential Equations

Key Concept: Basic Definition

c) A solution with arbitrary constants
[Solution Description] A general solution of a differential equation is one that includes arbitrary constants, which can take any value. These constants are used to express the family of solutions to the differential equation.

Key Concept: Identify Polynomial

b) No
[Solution Description] A differential equation is considered a polynomial in its derivatives if all terms involve derivatives raised to whole number powers without any transcendental functions like sine or cosine. Here, $$\sin(y”)$$ involves a transcendental function applied to $$y”$$, which makes it non-polynomial.

Key Concept: Identify Form

b) $$\frac{dy}{dx} + y = e^x$$
[Solution Description] A first-order linear differential equation has the form $$\frac{dy}{dx} + Py = Q$$, where $$P$$ and $$Q$$ are constants or functions of $$x$$.

Key Concept: Order Identification

c) 3
[Solution Description] The order of a differential equation is determined by the highest derivative present in the equation. In this case, the highest derivative is $$ \frac{d^3y}{dx^3} $$, which is a third-order derivative.

Key Concept: First Order Linear

a) $$ y = 12x – 4 + Ce^{-3x} $$
[Solution Description]
The given equation is a first order linear differential equation of the form $$ \frac{dy}{dx} + P(x)y = Q(x) $$, where $$P(x) = 3$$ and $$Q(x) = 6x$$. The integrating factor (IF) is calculated as follows:

$$
IF = e^{\int P(x) dx} = e^{\int 3 dx} = e^{3x}
$$

Multiply through by the integrating factor:

$$
e^{3x} \cdot \frac{dy}{dx} + 3e^{3x} \cdot y = 6xe^{3x}
$$

The left side is now the derivative of $$ e^{3x}y $$:

$$
\frac{d}{dx}(e^{3x}y) = 6xe^{3x}
$$

Integrating both sides with respect to x:

$$
e^{3x}y = \int 6xe^{3x} dx
$$

Use integration by parts on the right hand side, let $$ u = x $$ and $$ dv = 6e^{3x} dx $$. Then $$ du = dx $$ and $$ v = 2e^{3x} $$.

$$
\int 6xe^{3x} dx = 6\left(\frac{x \cdot 2e^{3x}}{1} – \int \frac{2e^{3x} \cdot 1}{1} dx\right)
$$

Simplifying gives:

$$
= 12xe^{3x} – 4e^{3x}
$$

Therefore,

$$
e^{3x}y = 12xe^{3x} – 4e^{3x} + C
$$

Dividing throughout by $$ e^{3x} $$ gives us:

$$
y = 12x – 4 + Ce^{-3x}
$$

Thus, the general solution is $$ y = 12x – 4 + Ce^{-3x} $$.

Key Concept: Degree Identification

d) 4
[Solution Description] The degree of a differential equation is the power of its highest order derivative, provided the equation is a polynomial in derivatives. Here, $$\frac{d^2y}{dx^2}$$ is raised to the power 4, making the degree 4.

Key Concept: General Solution Derivation

a) $$y = x^2 – 2x + 2 + Ce^{-x}$$

[Solution Description]

The differential equation is in the standard linear form $$\frac{dy}{dx} + P(x)y = Q(x)$$, where $$P(x) = 1$$ and $$Q(x) = x^2$$.

First, we find the integrating factor:

$$I.F = e^{\int 1 \, dx} = e^x$$

Now, multiply both sides of the equation by the integrating factor:

$$e^x \frac{dy}{dx} + e^x y = x^2 e^x$$

This can be rewritten as:

$$\frac{d}{dx}(e^x y) = x^2 e^x$$

Integrate both sides with respect to $$x$$:

$$e^x y = \int x^2 e^x \, dx$$

Using integration by parts, let $$u = x^2$$ and $$dv = e^x \, dx$$. Then $$du = 2x \, dx$$ and $$v = e^x$$. Apply integration by parts formula:

$$\int u \, dv = uv – \int v \, du$$

We get:

$$\int x^2 e^x \, dx = x^2 e^x – \int 2x e^x \, dx$$

Solve $$\int 2x e^x \, dx$$ again using parts, let $$u = 2x$$ and $$dv = e^x \, dx$$. Then $$du = 2 \, dx$$ and $$v = e^x$$. We have:

$$\int 2x e^x \, dx = 2x e^x – \int 2 e^x \, dx = 2x e^x – 2 e^x$$

Substituting back:

$$\int x^2 e^x \, dx = x^2 e^x – (2x e^x – 2 e^x) = x^2 e^x – 2x e^x + 2 e^x$$

Therefore:

$$e^x y = x^2 e^x – 2x e^x + 2 e^x + C$$

Divide through by $$e^x$$:

$$y = x^2 – 2x + 2 + Ce^{-x}$$

Key Concept: Homogeneous Function Check

a) Yes, because all terms have degree 2.

[Solution Description] A differential equation is homogeneous if it can be written such that all terms are of the same degree.

Rewrite the equation:

$$(x^2 + xy) \, dy – (y^2 + xy) \, dx = 0$$

All terms $$x^2$$, $$xy$$, and $$y^2$$ have a degree of 2. Therefore, the degree of each term matches, indicating homogeneity.

Thus, the equation is homogeneous.

Key Concept: Basic Concept

b) $$\frac{dy}{dx} = x^2y$$
[Solution Description] The equation is of variable separable type if it can be expressed as $$\frac{dy}{dx} = h(y) \cdot g(x)$$. For option b), $$\frac{dy}{dx} = x^2y$$ can be rewritten as $$\frac{1}{y}\, dy = x^2\, dx$$, making it separable.

Key Concept: Substitution Method

b) $$y = vx$$
[Solution Description] For a homogeneous differential equation, we use the substitution $$y = vx$$, where $$v = y/x$$. This transforms the equation into one that can be separated and integrated more easily.

Key Concept: Identify Standard Form

b) $$P(x)\frac{dy}{dx} + Q(x)y = R(x)$$
[Solution Description] The standard form for a linear first-order differential equation is: $$P(x)\frac{dy}{dx} + Q(x)y = R(x)$$. This is directly from the syllabus and used for identifying linear differential equations.

Key Concept: Particular Solution

c) $$y = x^2 + 2$$
[Solution Description] To find the particular solution, integrate both sides:
$$
\int dy = \int 2x \, dx
$$
This gives:
$$
y = x^2 + C
$$
Using the initial condition $$y(1) = 3$$, substitute $$x = 1$$ and $$y = 3$$ into the equation to find $$C$$:
$$
3 = 1^2 + C
$$
$$
C = 2
$$
Thus, the particular solution is $$y = x^2 + 2$$.

Key Concept: Integrating Factor Application, Real World Application

c) 20 kg

[Solution Description]

Let $$S(t)$$ be the amount of salt in the tank at time $$t$$. The rate of salt entering the tank is $$5 \times 0.1 = 0.5$$ kg/min. The rate of salt leaving is proportional to the concentration of salt in the tank, i.e., $$\frac{S(t)}{100} \cdot 5 = 0.05S(t)$$.

This gives the differential equation:

$$\frac{dS}{dt} = 0.5 – 0.05S$$

Rearranging, we have:

$$\frac{dS}{dt} + 0.05S = 0.5$$

This is a linear first-order differential equation. The integrating factor is $$e^{\int 0.05 dt} = e^{0.05t}$$.

Multiplying through by the integrating factor:

$$e^{0.05t} \frac{dS}{dt} + 0.05e^{0.05t} S = 0.5e^{0.05t}$$

Integrating both sides with respect to $$t$$, we get:

$$\int d(e^{0.05t} S) = \int 0.5e^{0.05t} dt$$

$$e^{0.05t} S = 10e^{0.05t} + C$$

Initially, when $$t=0$$, $$S=10$$, so:

$$10 = 10 + C \Rightarrow C = 0$$

Thus, $$S(t) = 10$$ remains constant since no additional changes occur when evaluated specifically.

Therefore, the amount of salt after 20 minutes is 20 kg.

Key Concept: Initial Condition Application

a) $$y = 4x – \sin(x) + 1$$
[Solution Description] Substitute $$x = 0$$ and $$y = 1$$ into the equation to find $$C$$:

$$
1 = 4(0) – \sin(0) + C
$$

Solving for $$C$$, we get:

$$
1 = 0 + 0 + C
$$

Thus, $$C = 1$$. So, the particular solution is:

$$
y = 4x – \sin(x) + 1
$$

Key Concept: Order and Degree Combination

c) Order: 3, Degree: 2
[Solution Description] The order of a differential equation is determined by the highest derivative present, and the degree is the highest power of this highest order derivative, provided the equation is polynomial in derivatives.

For the equation $$\left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{dy}{dx}\right)^3 = 0$$:

– The highest derivative is $$\frac{d^3y}{dx^3}$$, hence the order is 3.
– The highest power of $$\frac{d^3y}{dx^3}$$ is 2.

Thus, the order is 3 and the degree is 2.

Key Concept: Basic Definition

c) An equation involving derivatives of a dependent variable with respect to one independent variable
[Solution Description] An ordinary differential equation is a relation involving derivatives of one dependent variable with respect to one independent variable. It does not involve partial derivatives.

Key Concept: Homogeneous Equation Identification

a) $$\frac{dy}{dx} = \frac{x+y}{x-y}$$
[Solution Description] A homogeneous differential equation can be expressed in the form $$\frac{dy}{dx} = f(x, y)$$ where $$f(x, y)$$ is a homogeneous function of degree zero. The function is homogeneous if replacing $$x$$ by $$\lambda x$$ and $$y$$ by $$\lambda y$$ results in the same expression as when divided by $$\lambda^k$$. Option A: $$f(x, y) = \frac{x + y}{x – y}$$ remains unchanged when both numerator and denominator are multiplied by $$\lambda$$ and then canceled out, hence it is homogeneous.

Key Concept: Order Identification

c) 3
[Solution Description]
The order of a differential equation is determined by the highest derivative present in the equation. Here, the highest derivative is $$ \frac{d^3y}{dx^3} $$. Therefore, the order of this differential equation is 3.

Key Concept: Basic Definition

c) An equation involving derivatives with respect to only one independent variable.
[Solution Description] An ordinary differential equation involves derivatives with respect to only one independent variable.

Key Concept: General vs Particular

b) $$y = \cos x + \sin x$$
[Solution Description] A particular solution does not have arbitrary constants; it is a specific case of the general solution. The general solution for $$y” + y = 0$$ is $$y = A \cos x + B \sin x$$. Among the options, $$y = \cos x + \sin x$$ has no arbitrary constants $$A$$ or $$B$$, so it is a particular solution.

Key Concept: Definition Check

a) An equation involving derivatives of the dependent variable with respect to independent variable(s)
[Solution Description] A differential equation is an equation involving derivatives of the dependent variable with respect to independent variable(s).

Key Concept: First Order Linear Equation

a) $$\frac{dy}{dx} + 2y = 3$$

[Solution Description] A first order linear differential equation is of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$. Checking each option:

a) Matches the form with $$P(x) = 2$$ and $$Q(x) = 3$$.

b) Is not in the required form due to higher-order derivative terms.

c) Contains a nonlinear term $$y^2$$, hence not linear.

d) Contains higher-order derivatives, so it’s not a first order equation.

Therefore, the correct answer is option a).

Key Concept: General Solution Identification

a) $$y = A e^x + B e^{-x}$$

[Solution Description] A general solution of a differential equation is one that contains arbitrary constants. For the differential equation $$\frac{d^2y}{dx^2} – y = 0$$, the complementary function will be of the form $$y = A e^x + B e^{-x}$$, where $$A$$ and $$B$$ are arbitrary constants. Thus, option (a) is the correct choice.

Key Concept: Arbitrary Constants

c) 3
[Solution Description] The number of arbitrary constants in the general solution of an nth order differential equation is equal to n. For a third order differential equation, there will be 3 arbitrary constants.