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Popular >

y^'=e^{3x+4y}

Solution

y^{^{'}} = e^{3 x + 4 y}

Solution

y = - \frac{ln ( - \frac{4}{3} e ^{3 x} - c _{1} )}{4}

Solution steps

y^{'} (x) = e^{3 x + 4 y}

Solve separable ODE:

y = - \frac{ln ( - \frac{4}{3} e ^{3 x} - c _{1} )}{4}

y = - \frac{ln ( - \frac{4}{3} e ^{3 x} - c _{1} )}{4}

Graph

Popular Examples

(dx)/(dy)x^3+y^3=3

\frac{d x}{d y} x^{3} + y^{3} = 3

x*y^'=(1-y^2)^{1/2}

x \cdot y^{^{'}} = (1 - y^{2})^{\frac{1}{2}}

(dy)/(dx)=(1-13x)y^2

\frac{d y}{d x} = (1 - 13 x) y^{2}

(dr)/(ds)=0.09r

\frac{d r}{d s} = 0.09 r

(1+e^x)(dy}{dx}=\frac{e^x)/y

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{3x+4y} ?
The solution for y^'=e^{3x+4y} is y=-(ln(-4/3 e^{3x}-c_{1}))/4