What is the solution for y^'2y=3e^{4x}-4sin(4x),y(0)=-5 ?

The solution for y^'2y=3e^{4x}-4sin(4x),y(0)=-5 is y=-(sqrt(3e^{4x)+4cos(4x)+93})/2

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y^'2y=3e^{4x}-4sin(4x),y(0)=-5

y^{^{'}} 2 y = 3 e^{4 x} - 4 sin (4 x), y (0) = - 5

y = - \frac{3 e ^{4 x} + 4 cos ( 4 x ) + 93}{2}

Solution steps

y^{'} (x) 2 y = 3 e^{4 x} - 4 sin (4 x)

Solve separable ODE:

y = - \frac{3 e ^{4 x} + 4 cos ( 4 x ) + 93}{2}

y = - \frac{3 e ^{4 x} + 4 cos ( 4 x ) + 93}{2}

3 y \cdot d x - x \cdot d y = 0

y^{^{'}} (t) = \frac{t}{( 1 + y ^{2} )}, y (0) = 0

y d x + (x - 9) d y = 0, y (10) = 49

\frac{d y}{d t} = 12 - \frac{y}{50}

v y ln (v) d y - y (v d y + y d v) = 0

What is the solution for y^'2y=3e^{4x}-4sin(4x),y(0)=-5 ?
The solution for y^'2y=3e^{4x}-4sin(4x),y(0)=-5 is y=-(sqrt(3e^{4x)+4cos(4x)+93})/2