Question

The differential equation whose general solutions are y=Asinx+Bcosx is *

Answer 1

Answer:

y"+y =0

Step-by-step explanation:

Y=asinx+bcosx

Y'=acosx-bsinx

Y"=-asinx-bcosx=-(asinx+bcosx)

=-y

y"+y=0

Answer 2

Given:

General solution of required differential equation-

y=Asinx+Bcosx -------------------(1)

To Find:

Differential equation for given solution

Solution:

We know that,

• $\bf{\dfrac{dy}{dx}(sinx)=cosx}$

• $\bf{\dfrac{dy}{dx}(cosx)=-sinx}$

Since, this solution have two arbitrary constants, so we have to differentiate it twice

On differentiating (1) w.r.t. x, we get

$\longrightarrow\tt{\dfrac{dy}{dx}=Acosx+B(-sinx)}$

$\longrightarrow\tt{\dfrac{dy}{dx}=Acosx-Bsinx}------(2)$

On differentiating (2) w.r.t. x, we get

$\longrightarrow\tt{\dfrac{d^{2} y}{dx^{2} }=A(-sinx)-Bcosx}$

$\longrightarrow\tt{\dfrac{d^{2} y}{dx^{2} }=-Asinx-Bcosx}$

$\longrightarrow\tt{\dfrac{d^{2} y}{dx^{2} }=-(Asinx+Bcosx)}$

From (1), we get

$\longrightarrow\tt{\dfrac{d^{2} y}{dx^{2} }=-(y)}$

$\longrightarrow\tt{\dfrac{d^{2} y}{dx^{2} }=-y}$

$\longrightarrow\tt\pink{\dfrac{d^{2} y}{dx^{2}}+y=0}$

Hence, required differential equation is $\bf\green{\dfrac{d^{2} y}{dx^{2}}+y=0}$.