Solve the differential equations:− ★.(x sin x+cos x)(d^2 y/dx^2 ) −x cos x(dy/dx) + y cos x=0.


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Question Number 95119 by Mr.D.N. last updated on 27/May/20

$Solve the differential equations : -$ $★ . (x \sin x + \cos x) \frac{d^{2} y}{{dx}^{2}} - x \cos x \frac{dy}{dx} + y \cos x = 0 .$
	Answered by niroj last updated on 23/May/20
	$2. (x sin x + cos x )(d^2 y/dx^2 ) − x cos x (dy/dx) + y cos x=0 Sol^n : (d^2 y/dx^2 ) − ((x cos x)/(x sin x+cos x)) (dy/dx) + ((cos x)/(x sin x+cos x)) y=0 We know the form of variable coefficient with constant If Part of CF is known,then we can use of special cases: (d^2 y/dx^2 )+P(dy/dx)+Qy =R.. Let, P= − ((x cos x)/(x sinx+ cos x)) Q= ((cos x)/(x sin x+cos x)) and R=0 ∴P+Qx=0 −((x cos x)/(x sin x+cos x)) + ((x cos x)/(x sin x+cos x))=0 If P+Qx=0, then part of CF u= x y=uv y=xv....(i) Special case: (d^2 v/dx^2 )+(p+(2/u)(du/dx))(dv/dx)=(R/u) (d^2 v/dx^2 )+(−((x cos x)/(x sin x+cos x)) + (2/x))(dv/dx)= (0/x) Put , (dv/dx)=t ⇒ (d^2 v/dx^2 )= (dt/dx) (dt/dx) +((2/x) − ((x cos x)/(x sinx+cos x)))t=0 IF= e^(∫ ((2/x) − ((x cos x)/(x sinx+ cos x)))dx) = e^((2log x −log (x sin x +cos x)) = e^(log x^2 ) .e^(log (x sin x +cos x)^(−1) ) = x^(2 ) . (1/(x sin x +cos x)) t. (x^2 /(x sin x+cos x)) = ∫ (x^2 /(x sinx+cos x)).0 dx+C_1 t (x^2 /(x sinx+cos x)) = C_1 t = ((x sin x+ cos x)/x^2 )C_1 (dv/dx) = ((x sin x+cos x)/x^2 )C_1 ∫ dv = C_1 ∫ ((xsin x+cos x)/x^2 )dx (x sin x+cos x)(d/dx) sinx+xcosx−sinx =xcos x ∫dv = C_1 [ (xsinx+cos x)∫(1/x^2 )−∫{(xsinx +cos x)(d/dx)∫(1/x^2 )dx}dx v= C_1 [ −(x sinx+cos x)(1/x) +∫xcos x(1/x)dx +C_2 v= C_1 [−(xsin x+cos x)(1/x)+sinx]+C_2 put value of v in eq^n (i) y= xv =C_1 [−x(x sinx+cos x)(1/x)+x sin x]+C_2 x = C_1 [ −xsinx −cos x+ x sinx]+C_2 x =− C_1 cos x+C_2 x ∴ y = −C_1 cos x+C_2 x ∴ y Whis is required differential equation.$
	$2 . (x \sin x + \cos x) \frac{d^{2} y}{{dx}^{2}} - x \cos x \frac{dy}{dx} + y \cos x = 0$ ${Sol}^{n} :$ $\frac{d^{2} y}{{dx}^{2}} - \frac{x \cos x}{x \sin x + \cos x} \frac{dy}{dx} + \frac{\cos x}{x \sin x + \cos x} y = 0$ $We know the form of variable coefficient with constant If Part of CF is$ $known, then we can use of special cases :$ $\frac{d^{2} y}{{dx}^{2}} + P \frac{dy}{dx} + Qy = R . .$ $Let, P = - \frac{x \cos x}{x sinx + \cos x}$ $Q = \frac{\cos x}{x \sin x + \cos x} and R = 0$ $∴ P + Qx = 0$ $- \frac{x \cos x}{x \sin x + \cos x} + \frac{x \cos x}{x \sin x + \cos x} = 0$ $If P + Qx = 0, then part of CF u = x$ $y = uv$ $y = xv . . . . (i)$ $Special case : \frac{d^{2} v}{{dx}^{2}} + (p + \frac{2}{u} \frac{du}{dx}) \frac{dv}{dx} = \frac{R}{u}$ $\frac{d^{2} v}{{dx}^{2}} + (- \frac{x \cos x}{x \sin x + \cos x} + \frac{2}{x}) \frac{dv}{dx} = \frac{0}{x}$ $Put, \frac{dv}{dx} = t \Rightarrow \frac{d^{2} v}{{dx}^{2}} = \frac{dt}{dx}$ $\frac{dt}{dx} + (\frac{2}{x} - \frac{x \cos x}{x sinx + \cos x}) t = 0$ $IF = e^{\int (\frac{2}{x} - \frac{x \cos x}{x sinx + \cos x}) dx}$ $= e^{(2 \log x - \log (x \sin x + \cos x)}$ $= e^{\log x^{2}} . e^{\log {(x \sin x + \cos x)}^{- 1}}$ $= x^{2} . \frac{1}{x \sin x + \cos x}$ $t . \frac{x^{2}}{x \sin x + \cos x} = \int \frac{x^{2}}{x sinx + \cos x} . 0 dx + C_{1}$ $t \frac{x^{2}}{x sinx + \cos x} = C_{1}$ $t = \frac{x \sin x + \cos x}{x^{2}} C_{1}$ $\frac{dv}{dx} = \frac{x \sin x + \cos x}{x^{2}} C_{1}$ $\int dv = C_{1} \int \frac{xsin x + \cos x}{x^{2}} dx$ $(x \sin x + \cos x) \frac{d}{dx}$ $sinx + xcosx - sinx = xcos x$ $\int dv = C_{1} [(xsinx + \cos x) \int \frac{1}{x^{2}} - \int {(xsinx + \cos x) \frac{d}{dx} \int \frac{1}{x^{2}} dx} dx$ $v = C_{1} [- (x sinx + \cos x) \frac{1}{x} + \int xcos x \frac{1}{x} dx + C_{2}$ $v = C_{1} [- (xsin x + \cos x) \frac{1}{x} + sinx] + C_{2}$ $put value of v in {eq}^{n} (i)$ $y = xv$ $= C_{1} [- x (x sinx + \cos x) \frac{1}{x} + x \sin x] + C_{2} x$ $= C_{1} [- xsinx - \cos x + x sinx] + C_{2} x$ $= - C_{1} \cos x + C_{2} x$ $∴ y = - C_{1} \cos x + C_{2} x$ $∴ y Whis is required differential equation .$
	Commented by Mr.D.N. last updated on 23/May/20
	outstanding aspect ��
	Commented by i jagooll last updated on 23/May/20
	coll man. ��
	Commented by niroj last updated on 23/May/20
	��
	Commented by peter frank last updated on 24/May/20

	$thank you$
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