x 2 y ′ ′ − x y ′ + 4 y = c o s ( l o g x ) + x s i n ( l o g x ) x = e t , y ′ = e − t y t ′ , y ′ ′ = e − 2 t ( y t ′ ′ − y t ′ ) e 2 t e − 2 t ( y t ′ ′ − y t ′ ) − e t e − t y t ′ + 4 y = c o s t + e t s i n t y t ′ ′ − 2 y t ′ + 4 y = c o s t + e t s i n t x^2 y''-xy'+4y=cos(logx)+xsin(logx)\\
x=e^t, y'=e^{-t}y'_t , y''=e^{-2t}(y''_t-y'_t)\\
e^{2t}e^{-2t}(y''_t-y'_t)-e^te^{-t}y'_t +4y=cost+e^tsint\\
y''_t-2y'_t+4y=cost+e^tsint x 2 y ′′ − x y ′ + 4 y = cos ( l o gx ) + x s in ( l o gx ) x = e t , y ′ = e − t y t ′ , y ′′ = e − 2 t ( y t ′′ − y t ′ ) e 2 t e − 2 t ( y t ′′ − y t ′ ) − e t e − t y t ′ + 4 y = cos t + e t s in t y t ′′ − 2 y t ′ + 4 y = cos t + e t s in t
1 ) y t ′ ′ − 2 y t ′ + 4 y = 0 λ 2 − 2 λ + 4 = 0 D = 4 − 16 = − 12 = 12 i 2 λ 1 = 2 − 2 3 i 2 = 1 − 3 i λ 2 = 2 + 2 3 i 2 = 1 + 3 i y 1 = C 1 e t c o s 3 t + C 2 e t s i n 3 t 1) y''_t-2y'_t+4y=0\\
\lambda^2 -2\lambda+4=0\\
D=4-16=-12=12i^2\\
\lambda_1=\frac{2-2\sqrt3 i}{2}=1-\sqrt3 i\\
\lambda_2=\frac{2+2\sqrt3 i}{2}=1+\sqrt3 i\\
y_1=C_1e^tcos\sqrt3 t+C_2e^tsin\sqrt3 t 1 ) y t ′′ − 2 y t ′ + 4 y = 0 λ 2 − 2 λ + 4 = 0 D = 4 − 16 = − 12 = 12 i 2 λ 1 = 2 2 − 2 3 i = 1 − 3 i λ 2 = 2 2 + 2 3 i = 1 + 3 i y 1 = C 1 e t cos 3 t + C 2 e t s in 3 t
2 ) a ) f 1 ( t ) = c o s t y 2 = a c o s t + b s i n t y 2 ′ = − a s i n t + b c o s t y 2 ′ ′ = − a c o s t − b s i n t − a c o s t − b s i n t + 2 a s i n t − 2 b c o s t + 4 a c o s t + 4 b s i n t = c o s t c o s t : − a − 2 b + 4 a = 1 s i n t : − b + 2 a + 4 b = 0 3 a − 2 b = 1 2 a + 3 b = 0 a = 3 13 , b = − 2 13 y 2 = 3 13 c o s t − 2 13 s i n t 2) a) f_1(t)=cost\\
y_2=acost+bsint\\
y'_2=-asint+bcost\\
y''_2=-acost-bsint\\
-acost-bsint+2asint-2bcost+4acost+4bsint=cost\\
cost:-a-2b+4a=1\\
sint:-b+2a+4b=0\\
3a-2b=1\\
2a+3b=0\\
a=\frac{3}{13}, b=-\frac{2}{13}\\
y_2=\frac{3}{13}cost-\frac{2}{13}sint 2 ) a ) f 1 ( t ) = cos t y 2 = a cos t + b s in t y 2 ′ = − a s in t + b cos t y 2 ′′ = − a cos t − b s in t − a cos t − b s in t + 2 a s in t − 2 b cos t + 4 a cos t + 4 b s in t = cos t cos t : − a − 2 b + 4 a = 1 s in t : − b + 2 a + 4 b = 0 3 a − 2 b = 1 2 a + 3 b = 0 a = 13 3 , b = − 13 2 y 2 = 13 3 cos t − 13 2 s in t
b ) f 2 ( t ) = e t s i n t y 3 = a e t c o s t + b e t s i n t = e t ( a c o s t + b s i n t ) y 3 ′ = e t ( a c o s t + b s i n t ) + e t ( − a s i n t + b c o s t ) = = e t ( ( a + b ) c o s t + ( b − a ) s i n t ) y 3 ′ ′ = e t ( ( a + b ) c o s t + ( b − a ) s i n t ) + + e t ( − ( a + b ) s i n t + ( b − a ) c o s t ) = = e t ( 2 b c o s t − 2 a s i n t ) b) f_2(t)=e^tsint\\
y_3=ae^tcost+be^tsint=e^t(acost+bsint)\\
y'_3=e^t(acost+bsint)+e^t(-asint+bcost)=\\
=e^t((a+b)cost+(b-a)sint)\\
y''_3=e^t((a+b)cost+(b-a)sint)+\\
+e^t(-(a+b)sint+(b-a)cost)=\\
=e^t(2bcost-2asint) b ) f 2 ( t ) = e t s in t y 3 = a e t cos t + b e t s in t = e t ( a cos t + b s in t ) y 3 ′ = e t ( a cos t + b s in t ) + e t ( − a s in t + b cos t ) = = e t (( a + b ) cos t + ( b − a ) s in t ) y 3 ′′ = e t (( a + b ) cos t + ( b − a ) s in t ) + + e t ( − ( a + b ) s in t + ( b − a ) cos t ) = = e t ( 2 b cos t − 2 a s in t )
e t ( 2 b c o s t − 2 a s i n t ) − 2 e t ( ( a + b ) c o s t + ( b − a ) s i n t ) + + 4 e t ( a c o s t + b s i n t ) = e t s i n t c o s t : 2 b − 2 ( a + b ) + 4 a = 0 s i n t : − 2 a − 2 ( b − a ) + 4 b = 1 a = 0 b = 1 2 y 3 = 1 2 e t s i n t e^t(2bcost-2asint)-2e^t((a+b)cost+(b-a)sint)+\\
+4e^t(acost+bsint)=e^tsint\\
cost:2b-2(a+b)+4a=0\\
sint:-2a-2(b-a)+4b=1\\
a=0\\ b=\frac{1}{2}\\
y_3=\frac{1}{2}e^tsint\\ e t ( 2 b cos t − 2 a s in t ) − 2 e t (( a + b ) cos t + ( b − a ) s in t ) + + 4 e t ( a cos t + b s in t ) = e t s in t cos t : 2 b − 2 ( a + b ) + 4 a = 0 s in t : − 2 a − 2 ( b − a ) + 4 b = 1 a = 0 b = 2 1 y 3 = 2 1 e t s in t
y = y 1 + y 2 + y 3 = = C 1 e t c o s 3 t + C 2 e t s i n 3 t + + 3 13 c o s t − 2 13 s i n t + 1 2 e t s i n t y=y_1+y_2+y_3=\\
=C_1e^tcos\sqrt3 t+C_2e^tsin\sqrt3 t+\\
+\frac{3}{13}cost-\frac{2}{13}sint+\frac{1}{2}e^tsint y = y 1 + y 2 + y 3 = = C 1 e t cos 3 t + C 2 e t s in 3 t + + 13 3 cos t − 13 2 s in t + 2 1 e t s in t
Return to x x x
y = C 1 x c o s ( 3 l o g ∣ x ∣ ) + C 2 x s i n ( 3 l o g ∣ x ∣ ) + + 3 13 c o s ( l o g ∣ x ∣ ) − 2 13 s i n ( l o g ∣ x ∣ ) + 1 2 x s i n ( l o g ∣ x ∣ ) y=C_1xcos(\sqrt3 log|x|)+C_2xsin(\sqrt3 log|x|)+\\
+\frac{3}{13}cos(log|x|)-\frac{2}{13}sin(log|x|)+\frac{1}{2}xsin(log|x|) y = C 1 x cos ( 3 l o g ∣ x ∣ ) + C 2 x s in ( 3 l o g ∣ x ∣ ) + + 13 3 cos ( l o g ∣ x ∣ ) − 13 2 s in ( l o g ∣ x ∣ ) + 2 1 x s in ( l o g ∣ x ∣ )
#340153 on Dec 2023
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this helped me a lot...as same as my teacher explained me OwO tq