What is the solution for t^2(dx)/(dt)+3tx=t^4log_{10}(t)+1 ?

The solution for t^2(dx)/(dt)+3tx=t^4log_{10}(t)+1 is x=(integral of t^5log_{10}(t)+t)/(t^3)

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t^2(dx)/(dt)+3tx=t^4log_{10}(t)+1

t^{2} \frac{d x}{d t} + 3 t x = t^{4} lo g_{10} (t) + 1

x = \frac{\int t ^{5} lo g _{10} ( t ) + t d t}{t ^{3}}

Solution steps

t^{2} \frac{d x}{d t} + 3 t x = t^{4} lo g_{10} (t) + 1

Solve linear ODE:

x = \frac{\int t ^{5} lo g _{10} ( t ) + t d t}{t ^{3}}

x = \frac{\int t ^{5} lo g _{10} ( t ) + t d t}{t ^{3}}

\frac{1}{x + x ln ( x )}

f (x) = (4 x^{2} + 3) (4 x - 5)

\frac{d}{d x} (ln (25))

x \to 0 lim (\frac{( 8 ^{x} - 1 )}{x})

\int \frac{d}{1} (x) d x

What is the solution for t^2(dx)/(dt)+3tx=t^4log_{10}(t)+1 ?
The solution for t^2(dx)/(dt)+3tx=t^4log_{10}(t)+1 is x=(integral of t^5log_{10}(t)+t)/(t^3)