Differentiation (Part-38) | S N De

 

Form the differential equation by eliminating $\,A\,$ and $\,B\,$ (or $\,a\,$ or $\,b\,$) [$\,1-12\,$]: 

$\,1.~~ y=Ae^x+Be^{-x}$

Sol. $~~~~~~~y=Ae^x+Be^{-x}\\ \Rightarrow \frac{dy}{dx}=Ae^x-Be^{-x} \\ \therefore \frac{d^2y}{dx^2}=Ae^x+Be^{-x} \\ \Rightarrow \frac{d^2y}{dx^2}=y\,\,\text{(ans.)}$

$\,2.~~y=ax+bx^2$

Sol. $~~~~~y=ax+bx^2\rightarrow(1) \\ \Rightarrow y'=\frac{dy}{dx}=a+2bx\rightarrow(2) \\ \therefore y''=\frac{d^2y}{dx^2}=2b \rightarrow(3)$

Now from $\,(1),(2),(3)~$ we get,

$\,y=(y'-2bx)x+bx^2~~[\text{By (1),(2)}]\\~~~=[y'-y''x]x+\frac 12y''x^2~~[\text{By (3)}] \\ \Rightarrow 2y=2xy'-2x^2y''+x^2y'' \\ \Rightarrow 2y=2xy'-x^2y'' \\ \Rightarrow x^2y''-2xy'+2y=0\,\,\text{(ans.)}$

$\,3.~~y=Ae^x+Be^{-x}+x^2$

Sol. $~~~\,y=Ae^x+Be^{-x}+x^2 \\ \Rightarrow y-x^2=Ae^x+Be^{-x} \rightarrow(1)\\ \therefore y'-2x=Ae^x-Be^{-x} \\ \Rightarrow y''-2=Ae^x+Be^{-x} \\ \Rightarrow y''-2=y-x^2~~[\text{By (1)}] \\ \Rightarrow y''-y=2-x^2~~\text{(ans.)}$

$\,4.~~ax^2+by^2=1$

Sol. $~~~~ax^2+by^2=1 \\ \Rightarrow x^2+\frac{by^2}{a}=\frac 1a \\ \Rightarrow  2x+\left(\frac{2by}{a}\right)y'=0 \\ \Rightarrow  \frac ba=-\frac{x}{yy'}\\ \Rightarrow  \frac ab=-\frac{yy'}{x} ~~~\rightarrow(1)\\ \text{Diff. both sides of (1) w.r.t  } ~~x,  \\~~~~~~ 0=-\frac{x\left(y'^2+yy''\right)-yy'}{x^2} \\ \Rightarrow  x\left(y'^2+yy''\right)-yy'=0 \\ \Rightarrow yy'=x(y'^2+yy'')~~~\text{(ans.)}$

$\,5.~~y=a\tan^{-1}x+b$

Sol. $\,~~y=a\tan^{-1}x+b \\ \therefore y'=\frac{a}{1+x^2} \\ \Rightarrow (1+x^2)y'=a \\ \therefore  2xy'+(1+x^2)y''=0 \\ \Rightarrow  (1+x^2)y''+2xy'=0~~~\text{(ans.)}$

$\,6.~~y=(ax+b)e^{-2x}$

Sol. $~~~~~y=(ax+b)e^{-2x}\\ \Rightarrow ye^{2x}=ax+b \\ \therefore e^{2x}y'+2e^{2x}y=a \rightarrow(1)~~[\text{where}~~y'=\frac{dy}{dx}] \\ \text{Differentiating (1) w.r.t }~~x,\\ ~~~~2e^{2x}y'+e^{2x}y''+4e^{2x}y+2e^{2x}y'=0\\ \Rightarrow  e^{2x}[y''+4y'+4y]=0~~[\text{where}~~y''=\frac{d^2y}{dx^2}] \\ \Rightarrow y''+4y'+4y=0~~(\text{ans.})$

$\,7.~~y=A\sin mx+B \cos mx$

Sol. $~~y=A\sin mx+B \cos mx \\ \therefore y'=m(A\cos mx-B \sin mx) \\ \therefore y''=-m^2(A\sin mx+B \cos mx) \\ \Rightarrow y''=-m^2y \\ \therefore y''+m^2y=0~~\text{(ans.)}$

$\,8.~~y=ax+bx^3$

Sol. $~~~y=ax+bx^3 \rightarrow(1) \\ \therefore y'=a+3bx^2 \rightarrow(2) \\ \therefore y''=6bx \\ \Rightarrow b=\frac{1}{6x} \cdot y'' \rightarrow(3)$

From $\,(2),(3)~~$ we get,

$\,y'=a+3 \times \frac{1}{6x} .y'' \times x^2 \\ \Rightarrow y'=a+\frac x2 .y''\\ \Rightarrow a=y'-\frac x2 .y''\rightarrow(4)$

Now, from $\,(1),(3),(4)~~$ we get,

$\,y=(y'-\frac x2 .y'')x+(\frac{1}{6x} \cdot y'')x^3 \\ \Rightarrow 6y=6xy'-3x^2y''+x^2y'' \\ \therefore 6y=6xy'-2x^2y'' \\ \Rightarrow 3y=3xy'-x^2y'' \\ \Rightarrow x^2y''-3xy'+3y=0~~\text{(ans.)}$

$\,9.~~x=e^{-t}(a\cos t+b\sin t)$

Sol. $~~x=e^{-t}(a\cos t+b\sin t) \\ \therefore xe^t=a\cos t+b\sin t\rightarrow(1) \\ \therefore e^t\frac{dx}{dt}+xe^t=-a\sin t+b\cos t \\ \therefore e^t \frac{d^2x}{dt^2}+e^t \frac{dx}{dt}+xe^t+e^t\frac{dx}{dt}=-a\cos t-b\sin t \\ \Rightarrow e^t\frac{d^2x}{dt^2}+2e^t\frac{dx}{dt}+xe^t=-(a\cos t+b\sin t) \\ \Rightarrow \frac{d^2x}{dt^2}+2\frac{dx}{dt}+x=-e^{-t}(a\cos t+b\sin t)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-x \\ \Rightarrow \frac{d^2x}{dt^2}+2\frac{dx}{dt}+2x=0~~\text{(ans.)} $