[tex] \huge \red{ \boxed{ \green{question}}}[/tex]
[tex]a \frac{ {d}^{4} y}{ {dx}^{4} } + by \: = {ce}^{( \frac{ { - x}^{2} }{2}) } \\ [/tex]
[tex] \rule{500pt}{5pt}[/tex]
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Verified answer
Answer:
[tex](xy2−ex31)dx=x2ydy
x \sqrt{2} ydxdy−xy2=−ex31 −(1)
let y2=t
∴2ydy=dt
[/tex]
1.
[tex]→2dxx2dt−xt=−e−x31
dxdt−x2t=−x22e−x31
P=x−2,Q=−x32e−x31
I.f.=e∫p.dx
=e∫x−2dx
=e−2logx
=elogx−2
=x−2
[/tex]
Completer solution
[tex]t×x−2=∫x−2×−x22ex3−1dx+c
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+c
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u →x43dx=du
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u →x43dx=dut×x−2=−∫32eudu+c
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u →x43dx=dut×x−2=−∫32eudu+ct×x−2=−32eu+c
[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u →x43dx=dut×x−2=−∫32eudu+ct×x−2=−32eu+c[/tex]
Substitute the value of t and u
[tex]∴y2×x−2=−32ex3−1+c
[tex]∴y2×x−2=−32ex3−1+c2x2y2+31ex3−1=c
[tex]∴y2×x−2=−32ex3−1+c2x2y2+31ex3−1=c[/tex]
Step-by-step explanation:
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