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Popular >

3x^'+x=1

Solution

3 x^{^{'}} + x = 1

Solution

x = - e^{\frac{- t - c _{1}}{3}} + 1

Solution steps

3 x^{'} (t) + x = 1

Solve separable ODE:

x = - e^{\frac{- t - c _{1}}{3}} + 1

x = - e^{\frac{- t - c _{1}}{3}} + 1

Popular Examples

y^'=(8cos(8x))/(3+2y),y(0)=-1

y^{^{'}} = \frac{8 cos ( 8 x )}{3 + 2 y}, y (0) = - 1

t^9(dy)/(dt)+2t^8y=0

t^{9} \frac{d y}{d t} + 2 t^{8} y = 0

(dy)/(dx)=-1*1/((x+9)^2)*(y+1)

\frac{d y}{d x} = - 1 \cdot \frac{1}{( x + 9 ) ^{2}} \cdot (y + 1)

(dy)/(dx)=4+y^2

\frac{d y}{d x} = 4 + y^{2}

(dy)/(dx)=(1+y^2)/((1+x^2)xy)

\frac{d y}{d x} = \frac{1 + y ^{2}}{( 1 + x ^{2} ) x y}

Frequently Asked Questions (FAQ)

What is the solution for 3x^'+x=1 ?
The solution for 3x^'+x=1 is x=-e^{(-t-c_{1})/3}+1