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Differential Equations Ncert Solutions Maths class 12th 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 | Contributor-Level 10

4 months ago

(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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