ANSWER on Question #66659 – Math – Differential Equations
QUESTION
True/false. Justify. The normal form of the differential equation
y ′ ′ − 4 x y ′ + ( 4 x 2 − 1 ) y = − 3 e x 2 sin 2 x y'' - 4xy' + (4x^2 - 1)y = -3e^{x^2} \sin 2x y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = − 3 e x 2 sin 2 x
is
( d 2 v d x 2 ) + v = − 3 sin 2 x , where v = y e − x 2 \left(\frac{d^2 v}{dx^2}\right) + v = -3 \sin 2x, \quad \text{where} \quad v = ye^{-x^2} ( d x 2 d 2 v ) + v = − 3 sin 2 x , where v = y e − x 2 SOLUTION
v = y e − x 2 ↔ y = v e x 2 v = ye^{-x^2} \leftrightarrow y = ve^{x^2} v = y e − x 2 ↔ y = v e x 2
Then
y ′ = d d x ( v e x 2 ) = v ′ e x 2 + 2 x v e x 2 = ( v ′ + 2 x v ) e x 2 y' = \frac{d}{dx}(ve^{x^2}) = v'e^{x^2} + 2xve^{x^2} = (v' + 2xv)e^{x^2} y ′ = d x d ( v e x 2 ) = v ′ e x 2 + 2 xv e x 2 = ( v ′ + 2 xv ) e x 2 y ′ ′ = d d x ( y ′ ) = d d x ( ( v ′ + 2 x v ) e x 2 ) = e x 2 d d x ( v ′ + 2 x v ) + 2 x ( v ′ + 2 x v ) e x 2 = = ( v ′ ′ + 2 v + 2 x v ′ + 2 x v ′ + 4 x 2 v ) e x 2 = ( v ′ ′ + 2 v + 4 x v ′ + 4 x 2 v ) e x 2 \begin{array}{l}
y'' = \frac{d}{dx}(y') = \frac{d}{dx}\left((v' + 2xv)e^{x^2}\right) = e^{x^2} \frac{d}{dx}(v' + 2xv) + 2x(v' + 2xv)e^{x^2} = \\
= (v'' + 2v + 2xv' + 2xv' + 4x^2v)e^{x^2} = (v'' + 2v + 4xv' + 4x^2v)e^{x^2}
\end{array} y ′′ = d x d ( y ′ ) = d x d ( ( v ′ + 2 xv ) e x 2 ) = e x 2 d x d ( v ′ + 2 xv ) + 2 x ( v ′ + 2 xv ) e x 2 = = ( v ′′ + 2 v + 2 x v ′ + 2 x v ′ + 4 x 2 v ) e x 2 = ( v ′′ + 2 v + 4 x v ′ + 4 x 2 v ) e x 2
We substitute the obtained derivatives in the initial equation
y ′ ′ − 4 x y ′ + ( 4 x 2 − 1 ) y = = ( v ′ ′ + 2 v + 4 x v ′ + 4 x 2 v ) e x 2 − 4 x ( v ′ + 2 x v ) e x 2 + ( 4 x 2 − 1 ) v e x 2 = = ( v ′ ′ + 2 v + 4 x v ′ + 4 x 2 v − 4 x ( v ′ + 2 x v ) + ( 4 x 2 − 1 ) v ) e x 2 = = ( v ′ ′ + 2 v + 4 x v ′ + 4 x 2 v − 4 x v ′ − 8 x 2 v + 4 x 2 v − v ) e x 2 = = ( v ′ ′ + 2 v − v ⏟ v + 4 x v ′ − 4 x v ′ ⏟ 0 + 8 x 2 v − 8 x 2 v ⏟ 0 ) e x 2 = ( v ′ ′ + v ) e x 2 \begin{array}{l}
y'' - 4xy' + (4x^2 - 1)y = \\
= (v'' + 2v + 4xv' + 4x^2v)e^{x^2} - 4x(v' + 2xv)e^{x^2} + (4x^2 - 1)ve^{x^2} = \\
= (v'' + 2v + 4xv' + 4x^2v - 4x(v' + 2xv) + (4x^2 - 1)v)e^{x^2} = \\
= (v'' + 2v + 4xv' + 4x^2v - 4xv' - 8x^2v + 4x^2v - v)e^{x^2} = \\
= \left(v'' + \underbrace{2v - v}_{v} + \underbrace{4xv' - 4xv'}_{0} + \underbrace{8x^2v - 8x^2v}_{0}\right)e^{x^2} = (v'' + v)e^{x^2}
\end{array} y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = = ( v ′′ + 2 v + 4 x v ′ + 4 x 2 v ) e x 2 − 4 x ( v ′ + 2 xv ) e x 2 + ( 4 x 2 − 1 ) v e x 2 = = ( v ′′ + 2 v + 4 x v ′ + 4 x 2 v − 4 x ( v ′ + 2 xv ) + ( 4 x 2 − 1 ) v ) e x 2 = = ( v ′′ + 2 v + 4 x v ′ + 4 x 2 v − 4 x v ′ − 8 x 2 v + 4 x 2 v − v ) e x 2 = = ( v ′′ + v 2 v − v + 0 4 x v ′ − 4 x v ′ + 0 8 x 2 v − 8 x 2 v ) e x 2 = ( v ′′ + v ) e x 2
Hence
y ′ ′ − 4 x y ′ + ( 4 x 2 − 1 ) y = − 3 e x 2 sin 2 x ↔ ( v ′ ′ + v ) e x 2 = − 3 e x 2 sin 2 x y'' - 4xy' + (4x^2 - 1)y = -3e^{x^2} \sin 2x \leftrightarrow (v'' + v)e^{x^2} = -3e^{x^2} \sin 2x y ′′ − 4 x y ′ + ( 4 x 2 − 1 ) y = − 3 e x 2 sin 2 x ↔ ( v ′′ + v ) e x 2 = − 3 e x 2 sin 2 x v ′ ′ + v = − 3 sin 2 x ↔ d 2 v d x 2 + v = − 3 sin 2 x v'' + v = -3 \sin 2x \leftrightarrow \frac{d^2 v}{dx^2} + v = -3 \sin 2x v ′′ + v = − 3 sin 2 x ↔ d x 2 d 2 v + v = − 3 sin 2 x
ANSWER: TRUE.
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#340153 on Dec 2023