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Exercise 6.3 · Q30

Q.Verify that y=xmy=x^m is a solution of x2d2ydx2−mxdydx+my=0x^2\dfrac{d^2y}{dx^2}-mx\dfrac{dy}{dx}+my=0.

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Differentiate y=xmy=x^m twice: dydx=mxm−1\dfrac{dy}{dx}=mx^{m-1}, d2ydx2=m(m−1)xm−2\dfrac{d^2y}{dx^2}=m(m-1)x^{m-2}. Substitute into x2d2ydx2−mxdydx+myx^2\dfrac{d^2y}{dx^2}-mx\dfrac{dy}{dx}+my: x2⋅m(m−1)xm−2−mx⋅mxm−1+m⋅xm=m(m−1)xm−m2xm+mxm=xm[m2−m−m2+m]=0x^2\cdot m(m-1)x^{m-2}-mx\cdot mx^{m-1}+m\cdot x^m=m(m-1)x^m-m^2x^m+mx^m=x^m\big[m^2-m-m^2+m\big]=0. So th …

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