Show that $y = \frac{1}{x}\int {f(x){\text{d}}x}$ is a solution of the differential equation

$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0$.

Hence solve $x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = {x^{ - \frac{1}{2}}},{\text{ }}x > 0$, given that $y = 2$ when $x = 4$.

METHOD 1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{1}{{{x^2}}}\int {f(x){\text{d}}x + \frac{1}{x}f(x)}$     M1M1A1

$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0$     AG

Note:     M1 for use of product rule, M1 for use of the fundamental theorem of calculus, A1 for all correct.

METHOD 2

$x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x)$

$\frac{{{\text{d}}(xy)}}{{{\text{d}}x}} = f(x)$     (M1)

$xy = \int {f(x){\text{d}}x}$     M1A1

$y = \frac{1}{x}\int {f(x){\text{d}}x}$     AG

[3 marks]

$y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + c} \right)$     A1A1

Note:     A1 for correct expression apart from the constant, A1 for including the constant in the correct position.

attempt to use the boundary condition     M1

$c = 4$     A1

$y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + 4} \right)$     A1

Note:     Condone use of integrating factor.

[5 marks]

Total [8 marks]

This question allowed for several different approaches. The most common of these was the use of the integrating factor (even though that just took you in a circle). Other candidates substituted the solution into the differential equation and others multiplied the solution by $x$ and then used the product rule to obtain the differential equation. All these were acceptable.

This was a straightforward question. Some candidates failed to use the hint of ‘hence’, and worked from the beginning using the integrating factor. A surprising number made basic algebra errors such as putting the $+ c$ term in the wrong place and so not dividing it by $\chi$.