x\" (t) + 4 x(t) = t + 4 ; x(0) = 1 , x\'(0) = 0 Solution This is a limit of for
ID: 2973865 • Letter: X
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x" (t) + 4 x(t) = t + 4 ; x(0) = 1 , x'(0) = 0Explanation / Answer
This is a limit of form [ infinity / infinity ] so we need to apply L' Hospital rule. If u don't know that rule; it is like this Lim ( x --> a ) [ f (x) / g (x) ] = Lim ( x --> a ) [ f ' (x) / g ' (x) ] Where ( ' ) represent 1st order derivative with respect to x, note that quotient rule of differentiation is not to be applied here; while differentiating, just simple differentiation of individual functions with respect to x. So we get Lim ( x-->0) [ { d/dx { integral 0 to x ( t^2 / ( t^4 + 1 ) ) } } / { d/dx ( x^3 ) ] Now there is another formula for above thing, it's like If y = Integral ( a to h(x) ) f(t)dt [where a is cons., h(x) is a function of x] then dy/dh = f(t) { at t = h(x) } For eg. If y = Integral ( 1 to x^3 ) { dt / ( 1 + t^4 ) } dy/d(x^3) = ( 1 + t^4 ) { at t = x^3 ) ( applying the formula ) = ( 1 + ( x^3 )^4 ) [ putting t = x^3 ] = ( 1 + x^12 ) ------------------- ( 1 ) so dy/dx = [ dy/d(x^3) ] * [ d(x^3)/dx ] = ( 1 + x^12 ) * 3x^2 [ from 1, & diff. x^3 ] = 3x^2 / ( 1 + x^12 ) So, Now d/dx [ integral 0 to x { t^2 / ( t^4 + 1 ) } ] = { t^2 / ( t^4 + 1 ) } = { x^2 / ( x^4 + 1 ) } [ at t = x ] So the limit becomes Lim ( x --> 0 ) [ { x^2 / ( x^4 + 1 ) } / d/dx(x^3) ] = Lim ( x --> 0 ) [ { x^2 / ( x^4 + 1 ) } / 3x^2 ] = ( 1/ 3 ) * Lim ( x --> 0 ) [ { 1 / ( x^4 + 1 ) ] cancelling x^2 and taking 1/3 out. = ( 1/ 3 ) * Lim ( x --> 0 ) [ { 1 / ( 0^4 + 1 ) ] putting the value of x. = ( 1/3 ) * 1 = ( 1 / 3 ) Hope this helped.
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