1) x 2 y ′ ′ + x y ′ − y = x 2 e x x^2y''+xy'-y=x^2e^x x 2 y ′′ + x y ′ − y = x 2 e x
We will replace
x = e t , t = l n ∣ x ∣ y ′ = d y d x = y t ′ x t ′ = y t ′ e t = e − t y t ′ y ′ ′ = d y ′ d x = ( e − t ⋅ y t ′ ) ′ e t = y t ′ ′ e − t − e − t y t ′ e t = e − 2 t ( y t ′ ′ − y t ′ ) x=e^t, t=ln|x|\\
y'=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{y'_t}{e^t}=e^{-t}y'_t\\
y''=\frac{dy'}{dx}=\frac{(e^{-t}\cdot y'_{t})'}{e^t}=\frac{y''_{t}e^{-t}-e^{-t}y'_t}{e^t}=e^{-2t}(y''_t-y'_t) x = e t , t = l n ∣ x ∣ y ′ = d x d y = x t ′ y t ′ = e t y t ′ = e − t y t ′ y ′′ = d x d y ′ = e t ( e − t ⋅ y t ′ ) ′ = e t y t ′′ e − t − e − t y t ′ = e − 2 t ( y t ′′ − y t ′ )
then
e 2 t e − 2 t ( y ′ ′ − y ′ ) + e t e − t y ′ − y = e 2 t ⋅ e e t y ′ ′ − y ′ + y ′ − y = e 2 t ⋅ e e t y ′ ′ − y = e 2 t ⋅ e e t e^{2t}e^{-2t}(y''-y')+e^te^{-t}y'-y=e^{2t}\cdot e^{e^t}\\
y''-y'+y'-y=e^{2t}\cdot e^{e^t}\\
y''-y=e^{2t}\cdot e^{e^t} e 2 t e − 2 t ( y ′′ − y ′ ) + e t e − t y ′ − y = e 2 t ⋅ e e t y ′′ − y ′ + y ′ − y = e 2 t ⋅ e e t y ′′ − y = e 2 t ⋅ e e t
a)
y ′ ′ − y = 0 λ 2 − 1 = 0 λ 1 = 1 , λ 2 = − 1 y 1 = e t , y 2 = e − t y 0 = c 1 e t + c 2 e − t y''-y=0\\
\lambda^2-1=0\\
\lambda_1=1, \lambda_2=-1\\
y_1=e^t, y_2=e^{-t}\\
y_0=c_1e^t+c_2e^{-t} y ′′ − y = 0 λ 2 − 1 = 0 λ 1 = 1 , λ 2 = − 1 y 1 = e t , y 2 = e − t y 0 = c 1 e t + c 2 e − t
b)
y = c 1 ( t ) e t + c 2 ( t ) e − t y=c_1(t)e^t+c_2(t)e^{-t}\\ y = c 1 ( t ) e t + c 2 ( t ) e − t
{ c 1 ′ e t + c 2 ′ e − t = 0 c 1 ′ e t − c 2 ′ e − t = e 2 t e e t 2 c 1 ′ e t = e 2 t e e t c 1 ′ = 1 2 e t e e t c 1 = 1 2 ∫ e t e e t d t = 1 2 ∫ e e t d ( e t ) = = 1 2 e e t + k 1 \left\{\begin{matrix}
c'_1e^t+c'_2e^{-t}=0 \\
c'_1e^t-c'_2e^{-t}=e^{2t}e^{e^t}
\end{matrix}\right.\\
2c'_1e^t=e^{2t}e^{e^t}\\
c'_1=\frac{1}{2}e^{t}e^{e^t}\\
c_1=\frac{1}{2}\int e^{t}e^{e^t}dt=\frac{1}{2}\int e^{e^t}d(e^t)=\\
=\frac{1}{2}e^{e^t}+k_1 { c 1 ′ e t + c 2 ′ e − t = 0 c 1 ′ e t − c 2 ′ e − t = e 2 t e e t 2 c 1 ′ e t = e 2 t e e t c 1 ′ = 2 1 e t e e t c 1 = 2 1 ∫ e t e e t d t = 2 1 ∫ e e t d ( e t ) = = 2 1 e e t + k 1
2 c 2 ′ e − t = − e 2 t e e t c 2 ′ = − 1 2 e 3 t e e t c 2 = − 1 2 ∫ e 3 t e e t d t = − 1 2 ∫ e 2 t e e t d ( e t ) 2c'_2e^{-t}=-e^{2t}e^{e^t}\\
c'_2=-\frac{1}{2}e^{3t}e^{e^t}\\
c_2=-\frac{1}{2}\int e^{3t}e^{e^t}dt=-\frac{1}{2}\int e^{2t}e^{e^t}d(e^t) 2 c 2 ′ e − t = − e 2 t e e t c 2 ′ = − 2 1 e 3 t e e t c 2 = − 2 1 ∫ e 3 t e e t d t = − 2 1 ∫ e 2 t e e t d ( e t )
let u = e t u=e^t u = e t
∫ e 2 t e e t d e t = ∫ u 2 e u d u = u 2 e u − ∫ 2 u e u d u = = u 2 e u − 2 ( u e u − ∫ e u d u ) = u 2 e u − 2 u e u + 2 e u c 2 = − 1 2 ( u 2 e u − 2 u e u + 2 e u ) + k 2 = = − 1 2 ( e 2 t e e t − 2 e t e e t + 2 e e t ) + k 2 \int e^{2t}e^{e^t}de^t=\int u^2e^udu=u^2e^u-\int2ue^udu=\\
=u^2e^u-2(ue^u-\int e^udu)=u^2e^u-2ue^u+2e^u\\
c_2=-\frac{1}{2}(u^2e^u-2ue^u+2e^u)+k_2=\\
=-\frac{1}{2}(e^{2t}e^{e^t}-2e^te^{e^t}+2e^{e^t})+k_2 ∫ e 2 t e e t d e t = ∫ u 2 e u d u = u 2 e u − ∫ 2 u e u d u = = u 2 e u − 2 ( u e u − ∫ e u d u ) = u 2 e u − 2 u e u + 2 e u c 2 = − 2 1 ( u 2 e u − 2 u e u + 2 e u ) + k 2 = = − 2 1 ( e 2 t e e t − 2 e t e e t + 2 e e t ) + k 2
Then
y ( t ) = ( 1 2 e e t + k 1 ) e t + + ( − 1 2 ( e 2 t e e t − 2 e t e e t + 2 e e t ) + k 2 ) e − t y ( x ) = ( 1 2 e x + k 1 ) x + + ( − 1 2 ( x 2 e x − 2 x e x + 2 e x ) + k 2 ) 1 x y(t)=(\frac{1}{2}e^{e^t}+k_1)e^t+\\
+(-\frac{1}{2}(e^{2t}e^{e^t}-2e^te^{e^t}+2e^{e^t})+k_2)e^{-t}\\
y(x)=(\frac{1}{2}e^x+k_1)x+\\
+(-\frac{1}{2}(x^2e^x-2xe^x+2e^x)+k_2)\frac{1}{x}\\ y ( t ) = ( 2 1 e e t + k 1 ) e t + + ( − 2 1 ( e 2 t e e t − 2 e t e e t + 2 e e t ) + k 2 ) e − t y ( x ) = ( 2 1 e x + k 1 ) x + + ( − 2 1 ( x 2 e x − 2 x e x + 2 e x ) + k 2 ) x 1
2)
y ′ ′ + a 2 y = c o s e c a x = 1 s i n a x y''+a^2y=cosecax=\frac{1}{sinax} y ′′ + a 2 y = cosec a x = s ina x 1
a)
y ′ ′ + a 2 y = 0 λ 2 + a 2 = 0 λ 1 = a i , λ 2 = − a i y 1 = c o s a x , y 2 = s i n a x y 0 = c 1 c o s a x + c 2 s i n a x y''+a^2y=0\\
\lambda^2+a^2=0\\
\lambda_1=ai, \lambda_2=-ai\\
y_1=cosax, y_2=sinax\\
y_0=c_1cosax+c_2sinax y ′′ + a 2 y = 0 λ 2 + a 2 = 0 λ 1 = ai , λ 2 = − ai y 1 = cos a x , y 2 = s ina x y 0 = c 1 cos a x + c 2 s ina x
b)
y = c 1 ( x ) c o s a x + c 2 ( x ) s i n a x { c 1 ′ c o s a x + c 2 ′ s i n a x = 0 − a c 1 ′ s i n a x + a c 2 ′ c o s a x = 1 s i n a x y=c_{1}(x)cosax+c_2(x)sinax\\
\left\{\begin{matrix}
c'_1cosax+c'_2 sinax=0 \\
-a c'_1sinax+ac'_2 cosax=\frac{1}{sinax}
\end{matrix}\right. y = c 1 ( x ) cos a x + c 2 ( x ) s ina x { c 1 ′ cos a x + c 2 ′ s ina x = 0 − a c 1 ′ s ina x + a c 2 ′ cos a x = s ina x 1
Δ = ∣ c o s a x s i n a x − a s i n a x a c o s a x ∣ = a Δ 1 = ∣ 0 s i n a x 1 s i n a x a c o s a x ∣ = − 1 Δ 2 = ∣ c o s a x 0 − a s i n a x 1 s i n a x ∣ = c o s a x s i n a x c 1 ′ = Δ 1 Δ = − 1 a c 1 = ∫ − 1 a d x = − x a + k 1 c 2 ′ = Δ 2 Δ = 1 a ⋅ c o s a x s i n a x c 2 = ∫ 1 a ⋅ c o s a x s i n a x d x = 1 a 2 l n ∣ s i n a x ∣ + k 2 \Delta=\begin{vmatrix}
cos ax & sinax\\
-asinax & acosax
\end{vmatrix}=a\\
\Delta_1=\begin{vmatrix}
0 & sinax\\
\frac{1}{sinax} & acosax
\end{vmatrix}=-1\\
\Delta_2=\begin{vmatrix}
cos ax & 0\\
-asinax & \frac{1}{sinax}
\end{vmatrix}=\frac{cosax}{sinax}\\
c'_{1}=\frac{\Delta_1}{\Delta}=\frac{-1}{a}\\
c_{1}=\int\frac{-1}{a}dx=-\frac{x}{a}+k_1\\
c'_2=\frac{\Delta_2}{\Delta}=\frac{1}{a}\cdot\frac{cosax}{sinax}\\
c_2=\int\frac{1}{a}\cdot\frac{cosax}{sinax}dx=\frac{1}{a^2}ln|sinax|+k_2 Δ = ∣ ∣ cos a x − a s ina x s ina x a cos a x ∣ ∣ = a Δ 1 = ∣ ∣ 0 s ina x 1 s ina x a cos a x ∣ ∣ = − 1 Δ 2 = ∣ ∣ cos a x − a s ina x 0 s ina x 1 ∣ ∣ = s ina x cos a x c 1 ′ = Δ Δ 1 = a − 1 c 1 = ∫ a − 1 d x = − a x + k 1 c 2 ′ = Δ Δ 2 = a 1 ⋅ s ina x cos a x c 2 = ∫ a 1 ⋅ s ina x cos a x d x = a 2 1 l n ∣ s ina x ∣ + k 2
y = ( − x a + k 1 ) c o s a x + ( 1 a 2 l n ∣ s i n a x ∣ + k 2 ) s i n a x y=(-\frac{x}{a}+k_1)cosax+(\frac{1}{a^2}ln|sinax|+k_2)sinax y = ( − a x + k 1 ) cos a x + ( a 2 1 l n ∣ s ina x ∣ + k 2 ) s ina x
3) y ′ ′ − cot x y ′ − sin 2 x y = cos x − cos 3 x y''-\cot x y'- \sin^2x y=\cos x-\cos^3x y ′′ − cot x y ′ − sin 2 x y = cos x − cos 3 x
We will replace
cos x = t , x = arccos t , sin x = 1 − t 2 y ′ = d y d x = y t ′ x t ′ = y t ′ − 1 1 − t 2 = − 1 − t 2 y t ′ y ′ ′ = d y ′ d x = − y t ′ ′ 1 − t 2 − y t ′ 1 2 1 − t 2 ( − 2 t ) − 1 1 − t 2 = = y t ′ ′ ( 1 − t 2 ) − 2 t y t ′ \cos x=t, x=\arccos t, \sin x =\sqrt{1-t^2}\\
y'=\frac{dy}{dx}=\frac{y'_t}{x'_t}=\frac{y'_t}{-\frac{1}{\sqrt{1-t^2}}}=-\sqrt{1-t^2} y'_t\\
y''=\frac{dy'}{dx}=\frac{-y''_t\sqrt{1-t^2}-y'_t\frac{1}{2\sqrt{1-t^2}}(-2t)}{-\frac{1}{\sqrt{1-t^2}}}=\\
=y''_t(1-t^2)-2ty'_t cos x = t , x = arccos t , sin x = 1 − t 2 y ′ = d x d y = x t ′ y t ′ = − 1 − t 2 1 y t ′ = − 1 − t 2 y t ′ y ′′ = d x d y ′ = − 1 − t 2 1 − y t ′′ 1 − t 2 − y t ′ 2 1 − t 2 1 ( − 2 t ) = = y t ′′ ( 1 − t 2 ) − 2 t y t ′
input in equation
y ′ ′ ( 1 − t 2 ) − t y ′ − t 1 − t 2 ( − 1 − t 2 ) y ′ − − ( 1 − t 2 ) y = t − t 3 y ′ ′ ( 1 − t 2 ) − ( 1 − t 2 ) y = t ( 1 − t 2 ) y ′ ′ − y = t , y = y 0 + y 1 a ) y ′ ′ − y = 0 λ 2 − 1 = 0 , λ 1 = 1 , λ 2 = − 1 y 1 = e t , y 2 = e − t y 0 = c 1 e t + c 2 e − t b ) y 1 = a t + b y''(1-t^2)-ty'-\frac{t}{\sqrt{1-t^2}}(-\sqrt{1-t^2})y'-\\
-(1-t^2)y=t-t^3\\
y''(1-t^2)-(1-t^2)y=t(1-t^2)\\
y''-y=t,\\
y=y_0+y_1\\
a) y''-y=0\\
\lambda^2-1=0, \lambda_1=1, \lambda_2=-1\\
y_1=e^t, y_2=e^{-t}\\
y_0=c_1e^t+c_2e^{-t}\\
b) y_1=at+b y ′′ ( 1 − t 2 ) − t y ′ − 1 − t 2 t ( − 1 − t 2 ) y ′ − − ( 1 − t 2 ) y = t − t 3 y ′′ ( 1 − t 2 ) − ( 1 − t 2 ) y = t ( 1 − t 2 ) y ′′ − y = t , y = y 0 + y 1 a ) y ′′ − y = 0 λ 2 − 1 = 0 , λ 1 = 1 , λ 2 = − 1 y 1 = e t , y 2 = e − t y 0 = c 1 e t + c 2 e − t b ) y 1 = a t + b
input in equation
− a t − b = t ⟹ a = − 1 , b = 0 , y 1 = − t y = c 1 e t + c 2 e − t − t y = c 1 e cos x + c 2 e − cos x − cos x -at-b=t\implies a=-1, b=0,\\
y_1=-t\\
y=c_1e^t+c_2e^{-t}-t\\
y=c_1e^{\cos x}+c_2e^{-\cos x}-\cos x − a t − b = t ⟹ a = − 1 , b = 0 , y 1 = − t y = c 1 e t + c 2 e − t − t y = c 1 e c o s x + c 2 e − c o s x − cos x
#332078 on Oct 2022
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