CBSE Class 12 Maths Exemplar Chapter 9: Differential Equations NCERT Solutions

The given differential equation is:

$$\backslash frac\{dy\}\{dx\}\; =\; 2^\{y-x\}$$

Using the property $a^\{m-n\}\; =\; \backslash frac\{a^m\}\{a^n\}$, we can rewrite the equation as:

$$\backslash frac\{dy\}\{dx\}\; =\; \backslash frac\{2^y\}\{2^x\}$$

To solve this, we separate the variables:

$$\backslash frac\{dy\}\{2^y\}\; =\; \backslash frac\{dx\}\{2^x\}$$

Or, in terms of negative exponents:

$$2^\{-y\}\; dy\; =\; 2^\{-x\}\; dx$$

Now, we integrate both sides:

$$\backslash int\; 2^\{-y\}\; dy\; =\; \backslash int\; 2^\{-x\}\; dx$$

The integral of $a^\{kx\}$ is $\backslash frac\{a^\{kx\}\}\{k\; \backslash log\; a\}$. Applying this:

$$\backslash frac\{2^\{-y\}\}\{-1\; \backslash cdot\; \backslash log\; 2\}\; =\; \backslash frac\{2^\{-x\}\}\{-1\; \backslash cdot\; \backslash log\; 2\}\; +\; C$$

Multiplying by $-\backslash log\; 2$ to simplify:

$$2^\{-y\}\; =\; 2^\{-x\}\; -\; C\; \backslash log\; 2$$

Rearranging the terms:

$$2^\{-x\}\; -\; 2^\{-y\}\; =\; C\; \backslash log\; 2$$

Let $K\; =\; C\; \backslash log\; 2$, where K is an arbitrary constant.

$$2^\{-x\}\; -\; 2^\{-y\}\; =\; K$$

Thus, the general solution is $2^\{-x\}\; -\; 2^\{-y\}\; =\; K$.

The general equation of a line in a plane is given by $y\; =\; mx\; +\; c$, where $m$ is the slope and $c$ is the y-intercept. A non-vertical line implies that the slope $m$ is finite, so $m\; \backslash neq\; \backslash tan\; \backslash frac\{\backslash pi\}\{2\}$.

To find the differential equation, we need to eliminate the arbitrary constants $m$ and $c$. We differentiate the equation with respect to $x$ twice.

First differentiation:

$$\backslash frac\{dy\}\{dx\}\; =\; m$$

This equation shows that the first derivative represents the slope $m$.

Second differentiation:

$$\backslash frac\{d\}\{dx\}\backslash left(\backslash frac\{dy\}\{dx\}\backslash right)\; =\; \backslash frac\{d\}\{dx\}(m)$$

$$\backslash frac\{d^2y\}\{dx^2\}\; =\; 0$$

Since $m$ is a constant, its derivative is zero. This second-order differential equation $\backslash frac\{d^2y\}\{dx^2\}\; =\; 0$ represents the family of all non-vertical lines in a plane, as it contains no arbitrary constants.

The given differential equation is:

$$\backslash frac\{dy\}\{dx\}\; =\; e^\{-2y\}$$

We can separate the variables by moving the terms involving $y$ to one side and $x$ to the other:

$$\backslash frac\{dy\}\{e^\{-2y\}\}\; =\; dx$$

This can be written as:

$$e^\{2y\}\; dy\; =\; dx$$

Now, integrate both sides:

$$\backslash int\; e^\{2y\}\; dy\; =\; \backslash int\; dx$$

Applying the integration rule $\backslash int\; e^\{ax\}\; dx\; =\; \backslash frac\{e^\{ax\}\}\{a\}$:

$$\backslash frac\{e^\{2y\}\}\{2\}\; =\; x\; +\; C$$ \quad (i)

Here, $C$ is the constant of integration. We are given the initial condition that $y\; =\; 0$ when $x\; =\; 5$. Substitute these values into equation (i) to find $C$:

$$\backslash frac\{e^\{2(0)\}\}\{2\}\; =\; 5\; +\; C$$

$$\backslash frac\{e^0\}\{2\}\; =\; 5\; +\; C$$

$$\backslash frac\{1\}\{2\}\; =\; 5\; +\; C$$

Solving for $C$:

$$C\; =\; \backslash frac\{1\}\{2\}\; -\; 5\; =\; \backslash frac\{1\; -\; 10\}\{2\}\; =\; -\backslash frac\{9\}\{2\}$$

Now, substitute the value of $C$ back into equation (i):

$$\backslash frac\{e^\{2y\}\}\{2\}\; =\; x\; -\; \backslash frac\{9\}\{2\}$$

We need to find the value of $x$ when $y\; =\; 3$. Substitute $y\; =\; 3$ into the equation:

$$\backslash frac\{e^\{2(3)\}\}\{2\}\; =\; x\; -\; \backslash frac\{9\}\{2\}$$

$$\backslash frac\{e^6\}\{2\}\; =\; x\; -\; \backslash frac\{9\}\{2\}$$

Solve for $x$:

$$x\; =\; \backslash frac\{e^6\}\{2\}\; +\; \backslash frac\{9\}\{2\}$$

$$x\; =\; \backslash frac\{e^6\; +\; 9\}\{2\}$$

Therefore, when $y\; =\; 3$, the value of $x$ is $\backslash frac\{e^6\; +\; 9\}\{2\}$.

The given differential equation is:

$$(x^2\; -\; 1)\; \backslash frac\{dy\}\{dx\}\; +\; 2xy\; =\; \backslash frac\{1\}\{x^2\; -\; 1\}$$

To solve this, we first convert it into the standard form of a linear differential equation, which is $\backslash frac\{dy\}\{dx\}\; +\; Py\; =\; Q$. Divide the entire equation by $(x^2\; -\; 1)$ (assuming $x^2\; -\; 1\; \backslash neq\; 0$):

$$\backslash frac\{dy\}\{dx\}\; +\; \backslash frac\{2x\}\{x^2\; -\; 1\}\; y\; =\; \backslash frac\{1\}\{(x^2\; -\; 1)^2\}$$

This is a linear differential equation of the first order. By comparing it with the standard form $\backslash frac\{dy\}\{dx\}\; +\; Py\; =\; Q$, we identify:

$P\; =\; \backslash frac\{2x\}\{x^2\; -\; 1\}$

$Q\; =\; \backslash frac\{1\}\{(x^2\; -\; 1)^2\}$

Next, we calculate the integrating factor (IF), which is given by $IF\; =\; e^\{\backslash int\; P\; dx\}$:

$$\backslash mathsf\{IF\}\; =\; e^\{\backslash int\; \backslash left(\backslash frac\{2x\}\{x^2\; -\; 1\}\backslash right)\; dx\}$$

To evaluate the integral $\backslash int\; \backslash frac\{2x\}\{x^2\; -\; 1\}\; dx$, let $t\; =\; x^2\; -\; 1$. Then, $dt\; =\; 2x\; dx$. Substituting this into the integral:

$$\backslash int\; \backslash frac\{dt\}\{t\}\; =\; \backslash log\; |t|\; =\; \backslash log\; |x^2\; -\; 1|$$

So, the integrating factor is:

$$\backslash mathsf\{IF\}\; =\; e^\{\backslash log\; |x^2\; -\; 1|\}\; =\; |x^2\; -\; 1|$$

For simplicity, we can take $IF\; =\; x^2\; -\; 1$ (assuming $x^2\; -\; 1\; >\; 0$ or considering the absolute value later).

The general solution of a linear differential equation is given by $y\; \backslash cdot\; IF\; =\; \backslash int\; (Q\; \backslash cdot\; IF)\; dx\; +\; K$, where $K$ is the constant of integration.

$$y\; \backslash cdot\; (x^2\; -\; 1)\; =\; \backslash int\; \backslash left(\backslash frac\{1\}\{(x^2\; -\; 1)^2\}\; \backslash cdot\; (x^2\; -\; 1)\backslash right)\; dx\; +\; K$$

$$y\; \backslash cdot\; (x^2\; -\; 1)\; =\; \backslash int\; \backslash frac\{1\}\{x^2\; -\; 1\}\; dx\; +\; K$$

The integral $\backslash int\; \backslash frac\{1\}\{x^2\; -\; 1\}\; dx$ is a standard integral, which evaluates to $\backslash frac\{1\}\{2\}\; \backslash log\; \backslash left|\backslash frac\{x\; -\; 1\}\{x\; +\; 1\}\backslash right|$.

$$y\; \backslash cdot\; (x^2\; -\; 1)\; =\; \backslash frac\{1\}\{2\}\; \backslash log\; \backslash left|\backslash frac\{x\; -\; 1\}\{x\; +\; 1\}\backslash right|\; +\; K$$

Thus, the general solution is:

$$y\; (x^2\; -\; 1)\; =\; \backslash frac\{1\}\{2\}\; \backslash log\; \backslash left|\backslash frac\{x\; -\; 1\}\{x\; +\; 1\}\backslash right|\; +\; K$$

The given differential equation is:

$$\backslash frac\{dy\}\{dx\}\; +\; 2xy\; =\; y$$

To solve this, we first rearrange the terms to identify the type of differential equation. Move the $y$ term to the left side:

$$\backslash frac\{dy\}\{dx\}\; +\; 2xy\; -\; y\; =\; 0$$

Factor out $y$:

$$\backslash frac\{dy\}\{dx\}\; +\; (2x\; -\; 1)\; y\; =\; 0$$

This equation is in the form $\backslash frac\{dy\}\{dx\}\; +\; P(x)\; y\; =\; Q(x)$, where $P(x)\; =\; 2x\; -\; 1$ and $Q(x)\; =\; 0$. This is a linear differential equation of the first order.

Alternatively, we can see this is a separable equation if we rearrange it as:

$$\backslash frac\{dy\}\{dx\}\; =\; -\; (2x\; -\; 1)\; y$$

Separating the variables:

$$\backslash frac\{dy\}\{y\}\; =\; -\; (2x\; -\; 1)\; dx$$

Now, integrate both sides:

$$\backslash int\; \backslash frac\{dy\}\{y\}\; =\; \backslash int\; -\; (2x\; -\; 1)\; dx$$

The integral of $\backslash frac\{1\}\{y\}$ is $\backslash log\; |y|$, and the integral of $-(2x\; -\; 1)$ is $-\; (x^2\; -\; x)$.

$$\backslash log\; |y|\; =\; -\; (x^2\; -\; x)\; +\; C\_1$$

where $C\_1$ is the constant of integration.

Exponentiate both sides to solve for $y$:

$$|y|\; =\; e^\{-\; (x^2\; -\; x)\; +\; C\_1\}$$

$$|y|\; =\; e^\{C\_1\}\; e^\{-\; (x^2\; -\; x)\}$$

Let $C\; =\; \backslash pm\; e^\{C\_1\}$. Since $e^\{C\_1\}$ is always positive, $C$ can be any non-zero constant. If we allow $y=0$ as a solution (which it is, as $\backslash frac\{d(0)\}\{dx\}\; +\; (2x-1)(0)\; =\; 0$), then $C$ can be zero as well. Thus, $C$ is an arbitrary constant.

$$y\; =\; C\; e^\{-\; (x^2\; -\; x)\}$$

The general solution is $y\; =\; C\; e^\{x\; -\; x^2\}$.