2 find a particular solution of the differential equation y...

Transcript

00:01 These students here we have a question which states that let y of x let y of x y is equal to f of x satisfies the equation f of x is equal to e to the power minus x plus e to the power x of cause x minus 2 x minus integral of 0 to x x minus 1 and f prime of t d t so we have to find y satisfies one of these these equations so we are we are given four options so we have to find the solution of this function let's solve it and check out which of these is true you are given the function that f of x equals to e to the power minus x plus e to the power x of cause x minus 2x minus of integral of 0 to x 0 to x and let's simplify let's multiply our function with each term so we'll get so we've got x minus t so we'll multiply f prime of t d t with x we'll get 0 to x x of f prime of t d t and by multiplying with t will get plus of integral of 0 to x t f prime of t d t now let's write our equation as f of x will be equal to e to the power minus x plus e to the power x of course x minus 2x minus of x as here we've got x as are we'll treat as constant terms so write x into 0 to x of f prime of t d t and then we'll write it as plus integral of 0 to x of f prime of d t and then we'll write it as plus integral of 0 to x of f prime of d t so here we'll now integrate this will be our first and this will be your second term let's now write our equation as e to the power of minus x plus e to the power x of cross x minus 2 x minus of x so upon integration we'll get and putting the limits so we get f of x minus f of zero so these are our upper and lowered limits and then we'll write our equation as plus of f of t into 0 to x minus integral of 0 to x we'll put integration by parts on these two functions so upon putting these the integration by parts we'll get t f of t 0 to x minus integral of all integral of 0 x f f t d t d t so upon by putting integration by parts we've got these terms now by simplifying our function we'll get e to the power minus x plus e to the power x of cos x minus 2x minus of x x f of x plus of minus with minus will give us plus of x f of 0 and plus of x by putting the limit we'll get x f of x and minus of integral of 0 to x f of d t now we can write our function as so we'll further write our equation as which will be equal to e to the power minus x plus e to the power x of cross x minus of 2x minus of x integral of this term and cancel so we'll get minus of integral of 0 to x f prime of t d t now let's put that x equal to 0 let's put x equal to 0 in our equation so by putting x equal to 0 in this let's say this equation a so we'll get our f of x then will be equal to 2 so now we can write our function as f of x will be equal to e to the power minus x plus e to the power x cause of x minus 2x plus 2x minus 2x minus of integral of 0 to x f t t t and we know that plus 2 and minus 2x will cancel each other so we'll get f of x equals e to the power minus x plus e to the power x of cause x minus of integral of 0 to x f of t d t now we'll take the derivation the first derivation of f of our function f of x so upon taking derivation will apply the product rule upon applying the product rule you get first term as it is e to the power x as it is differentiation of cause x will be minus of sine x and plus a second term residue which is cause x and integration of differentiation of e to the power minus x will give us minus e to the power minus x plus into the power x will give us 8 to to power x and now you can write our by taking by applying integration will get minus of f of x we have got our limit as x and 0 so write minus of f of x to 0 so we have just applied libney's rule by libney's rule we have found this value so now you can write our f prime of x or function f prime of x as minus of sine x e to the power minus x plus e to the power x and plus of cause x minus of e to the power minus x plus e to the power x and function f of minus function f of x so we can just bring the function f of x on the left side so it will be f of x plus of f prime of x will be f prime of x will be equal to minus sine x e to the power minus x plus e to the power x and plus of cause x minus of e to the power minus x plus e to the power x so we can write our first derivative as d y upon d x divide upon d x and our function f of x as plus of y equals to minus of x e to the power minus x and minus of sine e to the power sine x e to the power x e to the power x and minus of e to the power minus x cause x and plus of e to the power x cause x so our d y divided by d x plus y equals to we can write it as minus of e to the power minus x by taking let's first take.

10:31 So by taking e to the power, e to the power x common, we'll get cause of x minus sine x.

10:47 And by taking minus of e to the power x common, we'll get cause x minus of sinex...

Question

2. Find a particular solution of the differential equation $$y'' + y = e^t + \sin(t)$$ (a) $$te^t + t \sin(t)$$ (b) $$e^t/2 - t \cos(t)/2$$ (c) $$e^t/2 + \sin(t)/2$$ (d) $$te^t/2 + \cos(t)$$

Question

2. Find a particular solution of the differential equation $$y'' + y = e^t + \sin(t)$$ (a) $$te^t + t \sin(t)$$ (b) $$e^t/2 - t \cos(t)/2$$ (c) $$e^t/2 + \sin(t)/2$$ (d) $$te^t/2 + \cos(t)$$

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