# How do I solve the differential equation?

(Use C as an arbitrary constant. Give your answer using the form k - C f(x), where f(x) has been simplified.)

xy + y' = 130 x

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• Anonymous
2 years ago

y' + x*y = 130x

This is linear differential equation therefore the answer is y = y_h + y_p; solve first for y_h

y' + x*y = 0 -> y' = -x*y -> y'/y = -x -> ln(y) = C - ((x^2)/2) -> y = e^[C - ((x^2)/2)] -> y = C*exp(-(x^2)/2);

Now solve for y_h by letting C -> C(x)

[d/dx C(x)*exp(-(x^2)/2)] + x*[C(x)*exp(-(x^2)/2)] = 130x;

C'(x)*exp(-(x^2)/2) + C(x)*(-x)*exp(-(x^2)/2)] + x*[C(x)*exp(-(x^2)/2)] = 130x;

C'(x)*exp(-(x^2)/2) = 130x -> C'(x) = 130x*exp((x^2)/2); C(x) = 130*exp((x^2)/2)

Final result is y_h = 130*exp((x^2)/2)*exp(-(x^2)/2) = 130

y = C*exp(-(x^2)/2) + 130

• rotchm
Lv 7
2 years ago

y' = dy/dx. From high school algebra, you can rewrite this as

dy/(130 - y) = x dx

Now the only calculus part: Just integrate both sides and this also generates your arbitrary constant C.

Highschool again: Isolate "y" if need be & rewrite as k - C f(x). Done!

The moral here is that 95% of advanced math (college & university) is just highschool math. If you remember this, you will be able to solve most probs in math on your own!