Which of the following options is the general solution of...

Which of the following options is the general solution of the differential equation\\ $\frac{dy}{dx} = \frac{y \cosh x}{y - 5}$?\\ $\circ \quad y - 5 \ln y = -\sinh x + c$\ $\circ \quad y = 5 \ln y - \sinh x + c$\ $\circ \quad y - 5 \ln y = \sinh x + c$\ $\circ \quad y = -5 \ln y - \sinh x + c$\ $\circ \quad y = -5 \ln y + \sinh x + c$

Transcript

00:01 Hi, here for the given question we are given differential equation 4y' ' -27y ' -7y is equal to cos h of 7x -1 upon 2 e to the power x.

00:17 Now here first of all we will find the complementary solution.

00:21 So here we will consider the lhs of the equation.

00:24 So here we know that this can be written as 4lambda square -27lambda -7 is equal to 0.

00:32 To simplify this quadratic equation we have lambda -7 let the 4lambda plus 1 equals to 0 which implies lambda is equal to 7 and lambda is equal to lambda2 -1.

00:45 Now here using this values we can say that the complementary solution can be written as y is equal to c e to the power 7x plus this is c1 and this is c2 e to the power minus 1 by 4 x.

01:03 Now here this is the complementary solution.

01:06 Further we need to calculate what is the particular solution.

01:09 So here for the particular solution we will use the variation of parameter matrix.

01:14 So here this is complementary solution.

01:17 For particular solution here we know that the value of y1 is equal to e to the power 7x, y2 is equal to 1 upon e to the power 1 by 4x.

01:33 Similarly y1 ' is equal to derivative will be 7 e to the power 7x and here y2 ' is equal to minus 1 by 4 e to the power 1 upon 4 times of x.

01:48 So here further if we take double derivative so we can observe that a0 of y ' is equal to 4 and here we have fx is equal to cos h of 7x minus 1 by 2 times e to the power x.

02:06 So here further using this value we can say that we have equation as e to the power 7x multiplied with c1 ' of x plus c2 ' of x divided by e to the power x by 4 equals to 0.

02:23 And here further this equation can be rewritten as 7 times e to the power 7x c1 ' of x plus minus c2 ' of x divided by e to the power x by 4.

02:39 Here we have 4 times and here this value is equal to cos of h of 7x minus 1 by 2 times e to the power x divided by.

02:52 So here further we need to find the value of this constant.

02:56 So here in order to calculate this constant we need to calculate the gram scale of the function.

03:00 Here gram scale w is equal to e to the power 7x 1 upon e to the power x by 4.

03:06 Here we have 7 times e to the power 7x and here minus 1 upon 4 times e to the power x by 4.

03:12 So here by calculating the gram scale we have this value equals to 29...

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