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  • If y = A e^ m x + B e^ n x, show that d ^2 y / dx ^2 - m+ n dy /dx + mny = 0

 Q14  If y = A e ^{mx} + Be ^{nx} , show that       \frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0                            

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best_answer

Given function is
y = A e ^{mx} + Be ^{nx}
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}                 -(i)
Now, second order derivative is 
\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}        -(ii)
Now, we need to show
\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0
Put the value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}  from equation (i) and (ii)
 m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})
 m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}+mnBe^{nx}
=0
Hence proved


 

Posted by

Gautam harsolia

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