Exercise 12.1
Question 7
यदि(If) y=Asinx+Bcosx, सिद्द करे कि (prove that) $\frac{d^{2} y}{d x^{2}}+y=0$Sol :
y=Asinx+Bcosx
Differentiating w.r.t x
$\frac{d y}{d x}=A \cos x+B(-\sin x)$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=A(-\sin x)-B\cos x$
$\frac{d^{2} y}{d x^{2}}=-\left[A \sin x+B\cos x\right]$
$\frac{d^{2} y}{d x^{2}}=-y$
$\frac{d^{2} y}{d x^{2}}+y=0$
Question 8
यदि(If) y=5cosx-3sinx , सिद्ध करे कि (prove that) $\frac{d^{2} y}{d x^{2}}+y=0$Sol :
y=5cosx-3sinx
Differentiating w.r.t x
$\frac{d y}{dx}=-5\sin x-3 \cos x$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-5 \cos x-3(-\sin x)$
$\frac{d^{2} y}{d x^{2}}=-[5\cos x-3 \sin x]$
$\frac{d^{2} y}{d x^{2}}=-y \Rightarrow \frac{d^{2} y}{d x^{2}}+y=0$
Question 9
A और B ज्ञात करे ताकि(Find A and B such that) y=Asin5x+Bcos5x समीकरण (satisfies the equation) $\frac{d^{2} y}{d x^{2}}+\frac{1}{5} \cdot \frac{d y}{d x}+15 y=101 \sin 5 x$ को संतुष्ट करता है ।Sol :
y=Asin5x+Bcos5x
Differentiating w.r.t x
$\frac{d y}{d x}=5 A \cos x-5B \sin 5 x$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-5 \times 5 A \sin 5 x-5 \times 5 B \cos 5 x$
=-25Asin5x-25Bcosx
∵$\frac{d^{2} y}{d x^{2}}+\frac{1}{5} \cdot \frac{d y}{d x}+15 y=101 \mathrm{sin} 5 x$
-25Asin5x-25Bcos5x+$\frac{1}{5}$(5Acos5x-5Bsin5x)+15(Asin5x+Bcos5x)=101sin5x
-25Asin5x-25Bcosx+Acos5x-Bsin5x+15Asin5x+15Bcos5x=101sin5x
-10Asin5x-Bsin5x-10Bcos5x+Acos5x=101sin5x
(-10A-B)sin5x+(-10B+A)cos5x=101sin5x+0.cos5x
By Equating
-10A-B=101..(i)×1
A-10B=0..(ii)×10
$\begin{aligned}-10A-B=101\\10A-100B=0 \\\hline -101B=101 \end{aligned}$
B=-1
समीकरण (ii) मे B का मान रखने पर ,
A-10(-1)=0
A=-10
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