1.
cos y cos ( p x ) + sin y sin ( p x ) = p \cos y\cos (px)+\sin y\sin (px)=p cos y cos ( p x ) + sin y sin ( p x ) = p
cos ( p x − y ) = p \cos(px-y)=p cos ( p x − y ) = p
p x − y = cos − 1 ( p ) px-y=\cos^{-1}(p) p x − y = cos − 1 ( p )
y = p x − cos − 1 ( p ) , p = d y d x y=px-\cos^{-1}(p), p=\dfrac{dy}{dx} y = p x − cos − 1 ( p ) , p = d x d y
This is a Clairaut differential equation.
Differentiating with respect to x , x, x , both sides
p = p + x d p d x + 1 1 − p 2 ⋅ d p d x p=p+x\dfrac{dp}{dx}+\dfrac{1}{\sqrt{1-p^2}}\cdot\dfrac{dp}{dx} p = p + x d x d p + 1 − p 2 1 ⋅ d x d p
( x + 1 1 − p 2 ) d p d x = 0 (x+\dfrac{1}{\sqrt{1-p^2}})\dfrac{dp}{dx}=0 ( x + 1 − p 2 1 ) d x d p = 0 Hence
x = − 1 1 − p 2 or d p d x = 0 = > p = c x=-\dfrac{1}{\sqrt{1-p^2}} \text{ or }\dfrac{dp}{dx}=0=>p=c x = − 1 − p 2 1 or d x d p = 0 => p = c
x = − 1 1 − p 2 = > 1 − p 2 = 1 x 2 , x < 0 x=-\dfrac{1}{\sqrt{1-p^2}}=>1-p^2=\dfrac{1}{x^2 }, \ x<0 x = − 1 − p 2 1 => 1 − p 2 = x 2 1 , x < 0
p 2 = x 2 − 1 x 2 p^2=\dfrac{x^2-1}{x^2} p 2 = x 2 x 2 − 1
p = ± x 2 − 1 x p=\pm\dfrac{\sqrt{x^2-1}}{x} p = ± x x 2 − 1 Substitute
y = ± ( x 2 − 1 − cos − 1 ( x 2 − 1 x ) ) , − 1 ≤ x < 0 y=\pm(\sqrt{x^2-1}-\cos^{-1}(\dfrac{\sqrt{x^2-1}}{x})), -1\leq x<0 y = ± ( x 2 − 1 − cos − 1 ( x x 2 − 1 )) , − 1 ≤ x < 0
p = d y d x = x x 2 − 1 − 1 x x 2 − 1 = x 2 − 1 x p=\dfrac{dy}{dx}=\dfrac{x}{\sqrt{x^2-1}}-\dfrac{1}{x\sqrt{x^2-1}}=\dfrac{\sqrt{x^2-1}}{x} p = d x d y = x 2 − 1 x − x x 2 − 1 1 = x x 2 − 1
Putting p = c p=c p = c in the equation we have
y = c x − cos − 1 ( c ) y=cx-\cos^{-1}(c) y = c x − cos − 1 ( c )
2.
( D 3 + 2 D 2 + D ) y = e 2 x + x 2 + sin ( 2 x ) (D^3+2D^2+D)y=e^{2x}+x^2+\sin(2x) ( D 3 + 2 D 2 + D ) y = e 2 x + x 2 + sin ( 2 x )
y ′ = z , y ′ ′ = z ′ , y ′ ′ ′ = z ′ ′ y'=z, y''=z', y'''=z'' y ′ = z , y ′′ = z ′ , y ′′′ = z ′′
z ′ ′ + 2 z ′ + z = e 2 x + x 2 + sin ( 2 x ) z''+2z'+z=e^{2x}+x^2+\sin(2x) z ′′ + 2 z ′ + z = e 2 x + x 2 + sin ( 2 x ) Homogeneous Equation
z ′ ′ + 2 z ′ + z = 0 z''+2z'+z=0 z ′′ + 2 z ′ + z = 0 The characteristic (auxiliary) equation
r 2 + 2 r + 1 = 0 r^2+2r+1=0 r 2 + 2 r + 1 = 0
r 1 = r 2 = − 1 r_1=r_2=-1 r 1 = r 2 = − 1
z h = c 1 x e − x + c 2 e − x z_h=c_1xe^{-x}+c_2e^{-x} z h = c 1 x e − x + c 2 e − x
z p = A e 2 x + B sin ( 2 x ) + C cos ( 2 x ) z_p=Ae^{2x}+B\sin(2x)+C\cos(2x) z p = A e 2 x + B sin ( 2 x ) + C cos ( 2 x )
+ ( M x 2 + N x + Q ) +(Mx^2+Nx+Q) + ( M x 2 + N x + Q )
z p ′ = 2 A e 2 x + 2 B cos ( 2 x ) − 2 C sin ( 2 x ) z_p'=2Ae^{2x}+2B\cos(2x)-2C\sin(2x) z p ′ = 2 A e 2 x + 2 B cos ( 2 x ) − 2 C sin ( 2 x )
+ ( 2 M x + N ) +(2Mx+N) + ( 2 M x + N )
z p ′ ′ = 4 A e 2 x − 4 B sin ( 2 x ) − 4 C cos ( 2 x ) + 2 M z_p''=4Ae^{2x}-4B\sin(2x)-4C\cos(2x)+2M z p ′′ = 4 A e 2 x − 4 B sin ( 2 x ) − 4 C cos ( 2 x ) + 2 M
Then
4 A e 2 x − 4 B sin ( 2 x ) − 4 C cos ( 2 x ) + 2 M 4Ae^{2x}-4B\sin(2x)-4C\cos(2x)+2M 4 A e 2 x − 4 B sin ( 2 x ) − 4 C cos ( 2 x ) + 2 M
+ 4 A e 2 x + 4 B cos ( 2 x ) − 4 C sin ( 2 x ) + 4 M x + 2 N +4Ae^{2x}+4B\cos(2x)-4C\sin(2x)+4Mx+2N + 4 A e 2 x + 4 B cos ( 2 x ) − 4 C sin ( 2 x ) + 4 M x + 2 N
+ A e 2 x + B sin ( 2 x ) + C cos ( 2 x ) + M x 2 + N x + Q +Ae^{2x}+B\sin(2x)+C\cos(2x)+Mx^2+Nx+Q + A e 2 x + B sin ( 2 x ) + C cos ( 2 x ) + M x 2 + N x + Q
= e 2 x + x 2 + sin ( 2 x ) =e^{2x}+x^2+\sin(2x) = e 2 x + x 2 + sin ( 2 x )
A = 1 9 A=\dfrac{1}{9} A = 9 1
− 3 B − 4 C = 1 -3B-4C=1 − 3 B − 4 C = 1
4 B − 3 C = 0 4B-3C=0 4 B − 3 C = 0
M = 1 2 M=\dfrac{1}{2} M = 2 1
N = − 2 N=-2 N = − 2
Q = 3 Q=3 Q = 3
z p = 1 9 e 2 x − 3 25 sin ( 2 x ) − 4 25 cos ( 2 x ) z_p=\dfrac{1}{9}e^{2x}-\dfrac{3}{25}\sin(2x)-\dfrac{4}{25}\cos(2x) z p = 9 1 e 2 x − 25 3 sin ( 2 x ) − 25 4 cos ( 2 x )
+ 1 2 x 2 − 2 x + 3 +\dfrac{1}{2}x^2-2x+3 + 2 1 x 2 − 2 x + 3 Therefore
z ( x ) = c 1 x e − x + c 2 e − x + 1 9 e 2 x z(x)=c_1xe^{-x}+c_2e^{-x}+\dfrac{1}{9}e^{2x} z ( x ) = c 1 x e − x + c 2 e − x + 9 1 e 2 x
− 3 25 sin ( 2 x ) − 4 25 cos ( 2 x ) + 1 2 x 2 − 2 x + 3 -\dfrac{3}{25}\sin(2x)-\dfrac{4}{25}\cos(2x)+\dfrac{1}{2}x^2-2x+3 − 25 3 sin ( 2 x ) − 25 4 cos ( 2 x ) + 2 1 x 2 − 2 x + 3
∫ x e − x d x = − x e − x + ∫ e − x d x \int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx ∫ x e − x d x = − x e − x + ∫ e − x d x
= − x e − x − e − x + C =-xe^{-x}-e^{-x}+C = − x e − x − e − x + C
y ( x ) = c 5 + c 3 x e − x + c 4 e − x + 1 18 e 2 x y(x)=c_5+c_3xe^{-x}+c_4e^{-x}+\dfrac{1}{18}e^{2x} y ( x ) = c 5 + c 3 x e − x + c 4 e − x + 18 1 e 2 x
+ 3 50 cos ( 2 x ) − 4 50 sin ( 2 x ) + 1 6 x 3 − x 2 + 3 x +\dfrac{3}{50}\cos(2x)-\dfrac{4}{50}\sin(2x)+\dfrac{1}{6}x^3-x^2+3x + 50 3 cos ( 2 x ) − 50 4 sin ( 2 x ) + 6 1 x 3 − x 2 + 3 x
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