Question

$\huge \red{ \boxed{ \green{question}}}$

$a \frac{ {d}^{4} y}{ {dx}^{4} } + by \: = {ce}^{( \frac{ { - x}^{2} }{2}) }$

$\rule{500pt}{5pt}$


Please don't spam

Really appreciated if solved on paper or by using latex

Thank You!​

Answer 1

Answer:

$(xy2−ex31)dx=x2ydy</p><p></p><p></p><p>x \sqrt{2} ydxdy−xy2=−ex31    −(1)</p><p></p><p></p><p>let y2=t</p><p></p><p>∴2ydy=dt</p><p></p><p>$

1.

$→2dxx2dt−xt=−e−x31</p><p></p><p></p><p>dxdt−x2t=−x22e−x31</p><p></p><p>P=x−2,Q=−x32e−x31</p><p></p><p></p><p>I.f.=e∫p.dx</p><p></p><p>      =e∫x−2dx</p><p></p><p>      =e−2logx</p><p></p><p>         =elogx−2</p><p></p><p>          =x−2</p><p></p><p>$

Completer solution

$t×x−2=∫x−2×−x22ex3−1dx+c</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+c</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u         →x43dx=du</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u         →x43dx=dut×x−2=−∫32eudu+c</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u         →x43dx=dut×x−2=−∫32eudu+ct×x−2=−32eu+c</u></p><p></p><p><u>[tex]t×x−2=∫x−2×−x22ex3−1dx+ct×x−2=−∫x42ex3−1dx+cLetx3−1=u         →x43dx=dut×x−2=−∫32eudu+ct×x−2=−32eu+c$

Substitute the value of t and u

$∴y2×x−2=−32ex3−1+c</u></p><p></p><p><u>[tex]∴y2×x−2=−32ex3−1+c2x2y2+31ex3−1=c</u></p><p></p><p><u>[tex]∴y2×x−2=−32ex3−1+c2x2y2+31ex3−1=c$

Step-by-step explanation:

HOPE IT HELPS BRO☺

BUT I AM NOT SURE