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Question: If the solution of the differential equation \(x\dfrac{dy}{dx}+y=x{{e}^{x}}\) be \(xy={{e}^{x}}\varp...

If the solution of the differential equation xdydx+y=xexx\dfrac{dy}{dx}+y=x{{e}^{x}} be xy=exφ(x)+cxy={{e}^{x}}\varphi (x)+c. Then, φ(x)\varphi (x) is equal to?
(a) x + 1
(b) x – 1
(c) 1 – x
(d) x

Explanation

Solution

Hint: For solving this question, we use the basic concepts of first order linear differential equations. The formula for first order linear equation dydx\dfrac{dy}{dx} + P(x)y = Q(x) is given by –
y e∫P(x)dx{{e}^{\int{P(x)dx}}} = ∫(Q(x)e∫P(x)dx) dx\int{\left( Q(x){{e}^{\int{P(x)dx}}} \right)}\text{ }dx + c

Complete step by step answer:
Thus, to solve this problem, we first convert xdydx+y=xexx\dfrac{dy}{dx}+y=x{{e}^{x}} into format similar to dydx\dfrac{dy}{dx} + P(x)y = Q(x). This would be because we can then use the formula y e∫P(x)dx{{e}^{\int{P(x)dx}}} = ∫(Q(x)e∫P(x)dx) dx\int{\left( Q(x){{e}^{\int{P(x)dx}}} \right)}\text{ }dx + c , to get the required differential equation. Thus, in the given problem, we have,
xdydx+y=xexx\dfrac{dy}{dx}+y=x{{e}^{x}}
Thus, dividing LHS and RHS by x, we get,
dydx+yx=ex\dfrac{dy}{dx}+\dfrac{y}{x}={{e}^{x}}
Now, we are in the position to compare this equation to dydx\dfrac{dy}{dx} + P(x)y = Q(x). By comparing, we have,
P(x) = 1x\dfrac{1}{x}, Q(x) = ex{{e}^{x}}
Now, evaluating, e∫P(x)dx{{e}^{\int{P(x)dx}}} , we have,
e∫1xdx{{e}^{\int{\dfrac{1}{x}dx}}}=eln⁡x{{e}^{\ln x}} = x (since, ∫1xdx\int{\dfrac{1}{x}dx} = lnx )
Now, substituting this in y e∫P(x)dx{{e}^{\int{P(x)dx}}} = ∫(Q(x)e∫P(x)dx) dx\int{\left( Q(x){{e}^{\int{P(x)dx}}} \right)}\text{ }dx + c, we have,
yx=∫xexdx+cyx=\int{x{{e}^{x}}dx}+c -- (1)
Now, we use integration by parts to evaluate ∫xexdx\int{x{{e}^{x}}dx}. The formula for integration of parts is given by –
∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int{uvdx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}}dx
Where, in this case, u = x and v = ex{{e}^{x}} . Thus, we have,
= u∫vdx−∫(dudx∫vdx)dxu\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}dx
= x∫exdx−∫(d(x)dx∫exdx)dxx\int{{{e}^{x}}dx-\int{\left( \dfrac{d\left( x \right)}{dx}\int{{{e}^{x}}dx} \right)}}dx
= xex−∫(1)exdxx{{e}^{x}}-\int{\left( 1 \right){{e}^{x}}dx}
(Since, d(x)dx=1\dfrac{d(x)}{dx}=1 and ∫exdx=ex\int{{{e}^{x}}dx={{e}^{x}}} )
= xex−exx{{e}^{x}}-{{e}^{x}}
Now, substituting this in (1), we get,
yx = xex−exx{{e}^{x}}-{{e}^{x}} + c
xy = ex(x−1){{e}^{x}}(x-1)+ c
Now, comparing this with xy=exφ(x)+cxy={{e}^{x}}\varphi (x)+c , we can tell that φ(x)\varphi (x) = (x-1).
Hence, the correct option is (b) x-1.

Note: In case of first order linear differential equation problems, it is always useful to remember the formula for dydx\dfrac{dy}{dx} + P(x)y = Q(x), given by y e∫P(x)dx{{e}^{\int{P(x)dx}}} = ∫(Q(x)e∫P(x)dx) dx\int{\left( Q(x){{e}^{\int{P(x)dx}}} \right)}\text{ }dx + c . At times equation may not be in the desired format, at those times it is suggested to manipulate the original differential equation slightly as done in the problem till we get equation of the format dydx\dfrac{dy}{dx} + P(x)y = Q(x).