What is the solution for (dx)/(dt)=6*0.3-5*x/(100+t),x(0)=0 ?

The solution for (dx)/(dt)=6*0.3-5*x/(100+t),x(0)=0 is

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(dx)/(dt)=60.3-5x/(100+t),x(0)=0

\frac{d x}{d t} = 6 \cdot 0.3 - 5 \cdot \frac{x}{100 + t}, x (0) = 0

x = \frac{1.8 ( 0.16666 \dots t ^{6} + 100 t ^{5} + 25000 t ^{4} + 3333333.33333 \dots t ^{3} + 250000000 t ^{2} + 10000000000 t )}{t ^{5} + 500 t ^{4} + 100000 t ^{3} + 10000000 t ^{2} + 500000000 t + 10000000000}

Solution steps

\frac{d x}{d t} = 6 \cdot 0.3 - 5 \cdot \frac{x}{100 + t}

Solve linear ODE:

x = \frac{1.8 ( 0.16666 \dots t ^{6} + 100 t ^{5} + 25000 t ^{4} + 3333333.33333 \dots t ^{3} + 250000000 t ^{2} + 10000000000 t )}{t ^{5} + 500 t ^{4} + 100000 t ^{3} + 10000000 t ^{2} + 500000000 t + 10000000000}

x = \frac{1.8 ( 0.16666 \dots t ^{6} + 100 t ^{5} + 25000 t ^{4} + 3333333.33333 \dots t ^{3} + 250000000 t ^{2} + 10000000000 t )}{t ^{5} + 500 t ^{4} + 100000 t ^{3} + 10000000 t ^{2} + 500000000 t + 10000000000}

d er i v a t i v e f (x) = 2 \cdot 7^{2 x^{2} + 3}

\int \frac{( x ^{3} )}{x ^{2} + 25} d x

\frac{\partial}{\partial t} (s cos (t))

t a y l or x - x^{3}, - 2

\frac{\partial}{\partial y} (3 x z^{2} e^{y^{2}})

What is the solution for (dx)/(dt)=6*0.3-5*x/(100+t),x(0)=0 ?
The solution for (dx)/(dt)=6*0.3-5*x/(100+t),x(0)=0 is