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Ncert Solutions Maths class 12th

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Ncert Solutions Maths class 12th Differential Equations

99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

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Vishal Baghel

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$\Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y} . . . . . . . . . . (1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l} y = v x \\ \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (v x) \\ \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

Substituting the values of y and $\frac{d v}{d x}$ in equation (1), we get:

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{x^{3} - 3 x {(v x)}^{2}}{{(v x)}^{3} - 3 x^{2} (v x)} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v (v^{3} - 3 v)}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v} \\ \Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x} \end{array}$

Integrating both sides, we get:

$\begin{array}{l} \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = l o g x + l o g C^{'} . . . . . . . . . (2) \\ N o w, \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \int \frac{v^{3} d v}{1 - v^{4}} - 3 \int \frac{v d v}{1 - v^{4}} \\ \Rightarrow \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = I_{1} - 3 I_{2}, W h e r e, I_{1} = \int \frac{v^{3} d v}{1 - v^{4}} a n d I_{2} = \int \frac{v d v}{1 - v^{4}} . . . . . . . . . . . (3) \end{array}$

$\begin{array}{l} L e t, 1 - v^{4} = t . \\ ∴ \frac{d}{d v} (1 - v^{4}) = \frac{d t}{d v} \\ \Rightarrow - 4 v^{3} = \frac{d t}{d v} \\ \Rightarrow v^{3} d v = - \frac{d t}{4} \\ N o w, I_{1} = \int - \frac{d t}{4} = - l o g t = - \frac{1}{4} l o g (1 - v^{4}) \end{array}$

$\begin{array}{l} A n d, I_{2} = \int \frac{v d v}{1 - v^{4}} = \int \frac{v d v}{1 - (v^{2})^{2}} \\ L e t, v^{2} = p . \\ ∴ \frac{d}{d v} (v^{2}) = \frac{d p}{d v} \\ \Rightarrow 2 v = \frac{d p}{d v} \\ \Rightarrow v d v = \frac{p}{2} \\ \Rightarrow I_{2} = \frac{1}{2} \int \frac{d p}{1 - p^{2}} = \frac{1}{2 * 2} l o g | \frac{1 + p}{1 - p} | = \frac{1}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | \end{array}$

Substituting the values of $I_{1}$ and $I_{2}$ in equation (3), we get:

$\int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = - \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} |$

Therefore, equation (2) becomes:

$\begin{array}{l} \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | = l o g x + l o g C^{'} \\ \Rightarrow - \frac{1}{4} l o g [(1 - v^{4}) (\frac{1 + v^{2}}{1 - v^{2}})] = l o g C^{'} x \\ \Rightarrow \frac{{(1 + v^{2})}^{4}}{{(1 - v^{2})}^{2}} = {(C^{'} x)}^{- 4} \\ \Rightarrow \frac{{(1 + \frac{y^{2}}{x^{2}})}^{4}}{{(1 - \frac{y^{2}}{x^{2}})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow {(x^{2} - y^{2})}^{2} = C^{' 4} {(x^{2} + y^{2})}^{4} \\ \Rightarrow (x^{2} - y^{2}) = C^{' 2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C {(x^{2} + y^{2})}^{2}, w h e r e C = C^{' 2} \end{array}$

Hence, the given result is proved.

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Ncert Solutions Maths class 12th Differential Equations

98. Form the differential equation representing the family of curves ${(x - a)}^{2} + 2 y^{2} = a^{2}$ where a an arbitrary constant.

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Vishal Baghel

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Equation of the given family of curves is ${(x - a)}^{2} + 2 y^{2} = a^{2}$

$\begin{array}{l} {(x - a)}^{2} + 2 y^{2} = a^{2} \\ \Rightarrow x^{2} + a^{2} - 2 a x + 2 y^{2} = a^{2} \\ \Rightarrow 2 y^{2} = 2 a x - x^{2} . . . . . . . . . . (1) \end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l} 2 y \frac{d y}{d x} = \frac{2 a - 2 x}{2} \\ \Rightarrow \frac{d y}{d x} = \frac{a - x}{2 y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 a - 2 x^{2}}{4 x y} . . . . . . . . . . (2) \end{array}$

From equation (*1), we get:

$2 a x = 2 y^{2} + x^{2}$

On substituting this value in equation (3), we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{2 y^{2} + x^{2} - 2 x^{2}}{4 x y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y} \end{array}$

Hence, the differential equation of the family of curves is given as $\frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y}$

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Ncert Solutions Maths class 12th Differential Equations

97. For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:

$\begin{array}{l} (i) y a e^{2} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0 \\ (i i) y = e^{x} (a c o s x + b s i n x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0 \\ (i i i) y = x s i n 3 x : \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x = 0 \\ (i v) x^{2} = 2 y^{2} l o g y : (x^{2} + y^{2}) \frac{d y}{d x} - x y = 0 \end{array}$

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Vishal Baghel

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(i) $y a e^{2} + b e^{- x} + x^{2}$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = a e^{x} - b e^{- x} + 2 x$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S

$\begin{array}{l} x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 \\ = x (a e^{x} - b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 \\ = (x a e^{x} - b x e^{- x} + 2 x) + (2 a e^{x} - 2 b e^{- x} + 4 x) - (a x e^{x} + b x e^{- x} + x^{3}) + x^{2} - 2 \\ = 2 a e^{x} - 2 b e^{- x} + x^{2} + 6 x - 2 \\ \neq 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii) $y = e^{x} (a c o s x + b s i n x) = a e^{x} c o s x + b e^{x} s i n x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a . \frac{d}{d x} (e^{x} c o s x) + b . \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d y}{d x} = a (e^{x} c o s x - e^{x} s i n x) + b . (e^{x} s i n x + e^{x} c o s x) \\ \Rightarrow \frac{d y}{d x} = (a + b) e^{x} c o s x + (b - a) e^{x} s i n x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = (a + b) . \frac{d}{d x} (e^{x} c o s x) (b - a) \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = (a + b) . [e^{x} c o s x - e^{x} s i n x] + (b - a) [e^{x} s i n x + e^{x} c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [(a + b) (c o s x - s i n x) + (b - a) (s i n x + c o s x)] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [a c o s x - a s i n x + b c o s x - b s i n x + b s i n x + b c o s x - a s i n x - a c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = [2 e^{x} (b c o s x - a s i n x)] \end{array}$

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y \\ = 2 e^{x} (b c o s x - a s i n x) - 2 e^{x} [(a + b) c o s x + (b - a) s i n x] + 2 e^{x} (a c o s x + b s i n x) \\ = e^{x} [(2 b c o s x - 2 a s i n x) - (2 a c o s x + 2 b c o s x) - (2 b s i n x - 2 a s i n x) + (2 a c o s x + 2 b s i n x)] \\ = e^{x} [(2 b - 2 a - 2 b + 2 a) c o s x] + e^{x} [(- 2 a - 2 b + 2 a + 2 b) s i n x] \\ = 0 \end{array}$

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii)&nb

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(i) $y a e^{2} + b e^{- x} + x^{2}$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = a e^{x} - b e^{- x} + 2 x$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii) $y = e^{x} (a c o s x + b s i n x) = a e^{x} c o s x + b e^{x} s i n x$

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S of the given differential equation, we get:

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii) $y = x s i n 3 x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x s i n 3 x) = s i n 3 x + x . c o s 3 x . 3 \\ \Rightarrow \frac{d y}{d x} = s i n 3 x + 3 x c o s 3 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (s i n 3 x) + 3 \frac{d}{d x} (x c o s 3 x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 3 c o s 3 x + 3 [c o s 3 x + x (- s i n 3 x) . 3] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} 6 c o s 3 x - 9 x s i n 3 x \end{array}$

Substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x \\ = (6 . c o s 3 x - 9 x s i n 3 x) + 9 x s i n 3 x - 6 c o s 3 x \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

(iv) $x^{2} = 2 y^{2} l o g y$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} 2 x = 2 . \frac{d}{d x} = [y^{2} l o g y] \\ \Rightarrow x = [2 y . l o g y . \frac{d y}{d x} + y^{2} . \frac{1}{y} . \frac{d y}{d x}] \\ \Rightarrow x = \frac{d y}{d x} (2 y l o g y + y) \\ \Rightarrow \frac{d y}{d x} = \frac{x}{y (1 + 2 l o g y)} \end{array}$

Substituting the value of $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} (x^{2} + y^{2}) \frac{d y}{d x} - x y \\ = (2 y^{2} l o g y + y^{2}) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = y^{2} (1 + 2 l o g y) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = x y - x y \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

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Ncert Solutions Maths class 12th Differential Equations

96. For each of the differential equations given below, indicate its order and degree (if defined):

$\begin{array}{l} (i) \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x \\ (i i) {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x \\ (i i i) \frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0 \end{array}$

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Vishal Baghel

Contributor-Level 10

(i) Given: Differential equation $\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x$

The highest order derivative present in this differential equation is $\frac{d^{2} y}{d x^{2}}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation ${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivativ

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(i) Given: Differential equation $\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x$

The highest order derivative present in this differential equation is $\frac{d^{2} y}{d x^{2}}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation ${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d y}{d x}$ is 3.

Therefore, Order = 1, Degree = 3

(iii) Given: Differential equation $\frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0$

The highest order derivative present in this differential equation is $\frac{d^{4} y}{d x^{4}}$ and hence order of this differential equation if 4.

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

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Ncert Solutions Maths class 12th Differential Equations

95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$(1 - y^{2}) \frac{d x}{d y} + y x = \propto y (- 1 < y < 1)$

$\frac{d x}{d y} + \frac{y}{1 - y^{2}} * x = \frac{\propto y}{1 - y^{2}}$ which is of form

$\begin{array}{l} \frac{d x}{d y} + P x = Q & P = \frac{y}{1 - y^{2}} & Q = \frac{\propto y}{1 - y^{2}} \\ \int p d x = \int \frac{y}{1 - y^{2}} d x = - \frac{1}{2} \int \frac{- 2 y}{1 - y^{2} d x} \end{array}$

$\begin{array}{l} = - \frac{1}{2} l o g | 1 - y^{2} | \\ = l o g {[1 - y^{2}]}^{- \frac{1}{2}} \end{array}$

$∴ I . F = e^{\int P d x} = e^{l o g {[1 - y^{2}]}^{- \frac{1}{2}}} = {[1 - y^{2}]}^{- \frac{1}{2}}$

$∴$ option (D ) is correct.

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Ncert Solutions Maths class 12th Differential Equations

94. Choose the correct answer:

The integrating factor of the differential equation $\frac{d y}{d x} - y = 2 x^{2}$ is:

(A) e^–x

(B) e^–y

(C) $\frac{1}{x}$

(D) x

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Vishal Baghel

Contributor-Level 10

The given D.E is

$\begin{array}{l} x \frac{d y}{d x} - y = 2 x^{2} \\ \frac{d y}{d x} - \frac{1}{x} y = 2 x \end{array}$

Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = - \frac{1}{x}$

$∴ I . E = e^{\int P d x} = e^{\int - \frac{1}{x} d x} = e^{l o g - x} = e^{l o g - x^{- 1}} = x^{- 1} = \frac{1}{x}$

$∴$ Option (c) is correct

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Ncert Solutions Maths class 12th Differential Equations

93. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

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Vishal Baghel

Contributor-Level 10

We know that slope of tangent to the curve is $\frac{d y}{d x}$

$∴ x + y = \frac{d y}{d x} + 5$

$\frac{d y}{d x} - y = x - 5$ Which has form $\frac{d y}{d x} + P x = Q$

$w h e r e, P = 1 & Q = x - 5$

$∴ I . F = e^{\int P d x} = e^{\int - 1 d x} = e^{- x}$

Thus the solution has the form

$\begin{array}{l} y e^{- x} = \int (x - 5) e^{- x} d x + c \\ = \int x e^{- x} d x - \int 5 e^{- x} d x + c \\ = y e^{- x} = I + 5 e^{- x} + c \\ w h e r e, I = \int x e^{- x} d x \\ = x \int e^{- x} d x - \int \frac{d}{d x} x \int e^{- x} d x d x \\ = - x e^{- x} + \int e^{- x} d x \\ = - x e^{- x} - e^{- x} \end{array}$

$\begin{array}{l} ∴ y e^{- x} = - x e^{- x} - e^{- x} + 5 e^{- x} + c \\ = y e^{- x} = - x e^{- x} + 4 e^{- x} + c \\ = y = - x + 4 + \frac{c}{e^{- x}} \\ = y + x = 4 + c e^{x} \end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l} 2 + 0 = 4 c e^{0} \\ 2 - 4 = c \\ c = - 2 \end{array}$

$∴$ The particular solution is

$y + x = 4 - 2 e^{x}$

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Ncert Solutions Maths class 12th Differential Equations

92. Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of coordinates of that point.

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Vishal Baghel

Contributor-Level 10

We know the slope of tangent to curve is $\frac{d y}{d x}$ .

$∴$ $\frac{d y}{d x} = x + y$

$= \frac{d y}{d x} - y = x$ which has form $\frac{d y}{d x} + P y = Q$

So, $P = - 1 & Q = x$

$∴ I . F = e^{\int P d x} = e^{\int - d x} = e^{- x}$

Thus the solution is of the form .

$\begin{array}{l} y * e^{- x} = \int x . e^{- x} d x + c \\ = x \int e^{- x} d x - \int \frac{d x}{d x} \int e^{- x} d x d x + c \\ = - x e^{- x} + \int e^{- x} d x + c \\ = y e^{- x} = - x e^{- x} - e^{- x} + c \\ = y = - x - 1 + \frac{c}{e^{- x}} \\ = y + x + 1 = c e^{x} \end{array}$

Given, the curve passes through origin (0,0) i.e, $y = 0, w h e n, x = 0$

$0 + 0 + 1 = c e^{0} = c = 1$

$∴$ Thus, equation of the curve is

$y + x + 1 = e^{x}$

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Ncert Solutions Maths class 12th Differential Equations

91. $\frac{d y}{d x} - 3 y c o t x = s i n 2 x; y = 2 w h e n x = \frac{π}{2}$

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$\frac{d y}{d x} - 3 y c o t x = s i n 2 x$

$\frac{d y}{d x} - 3 c o t x . y = s i n 2 x$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = - 3 c o t x & Q = s i n 2 x$

$∴ I . F = e^{- \int P d x} = e^{\int - 3 c o t x d x} = e^{- 3 \int c o t x d x} = e^{- 3 l o g | c o t x d x |} = e^{l o g {(s i n)}^{- 3}} = \frac{1}{s i n^{3} x}$

Thus the solution is of the form.

$\begin{array}{l} y * \frac{1}{s i n^{3} x} = \int s i n 2 x . \frac{1}{s i n^{3} x} d x + c \\ = \int \frac{2 s i n x c o s x}{s i n^{3} x} d x + c {? s i n 2 x = 2 s i n x c o s x} \\ = \int 2 c o s e c x c o t x d x + c \\ = - 2 c o s e c x + c \\ = \frac{y}{s i n^{3} x} = \frac{- 2}{s i n x} + c \\ = - 2 y = - 2 s i n^{2} x + c s i n^{3} x \end{array}$

Given, $y = 2, w h e n, x = \frac{π}{2}$

$\begin{array}{l} 2 = - 2 s i n^{2} \frac{1}{2} + c s i n^{3} \frac{π}{2} \\ = 2 = - 2 + c \\ = e = 2 + 2 = 4 \end{array}$

$∴$ The particular solution is, $y = - 2 s i n^{2} x + 4 s i n^{3} x$

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Ncert Solutions Maths class 12th Differential Equations

90. $(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}; y = 0 w h e n x = 1$

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}$

$\frac{d y}{d x} + P x = Q$

So, $P = \frac{2 x}{1 + x^{2}} & Q = \frac{1}{{(1 + x^{2})}^{2}}$

$\int P d x = \int \frac{2 x}{1 + x^{2}} d x = l o g | 1 + x^{2} |$

$∴ I . F = e^{\int P d x} = e^{l o g | 1 + x^{2} |} = 1 + x^{2}$

Thus the solution is if form,

$y * (I . F) = \int Q . (I . F) d x + c$

$\begin{array}{l} y (1 + x^{2}) = \int \frac{1}{{(1 + x^{2})}^{2}} * (1 + x^{2}) d x + c \\ = \int \frac{1}{(1 + x^{2})} d x + c \\ y * (1 + x^{2}) = t a n^{- 1} x + c \end{array}$

Given, $y = 0, w h e n, x = 1$

$\begin{array}{l} 0 = t a n^{- 1} 1 + c \\ c = t a n^{- 1} 1 = - \frac{π}{4} \end{array}$

$∴$ The particular solution is

$y (1 + x^{2}) = t a n^{- 1} x - \frac{π}{4}$

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