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Popular >

xy^'-2y=x^4-4

Solution

x y^{^{'}} - 2 y = x^{4} - 4

Solution

y = \frac{x ^{4}}{2} + 2 + c_{1} x^{2}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{4}}{2} + 2 + c_{1} x^{2}

y = \frac{x ^{4}}{2} + 2 + c_{1} x^{2}

Popular Examples

((3y^2-t^2))/((y^5))*y^'+t/((2y^4))=0,y(1)=3

\frac{( 3 y ^{2} - t ^{2} )}{( y ^{5} )} \cdot y^{^{'}} + \frac{t}{( 2 y ^{4} )} = 0, y (1) = 3

derivative of (e^x-1)/x derivative of

\frac{e ^{x} - 1}{x}

integral of 1/(1-sin(θ)+cos(θ))

\int \frac{1}{1 - sin ( θ ) + cos ( θ )} d θ

integral of 3e^{4t}

\int 3 e^{4 t} d t

integral of (4+8xsin(x))/(2x)

\int \frac{4 + 8 x sin ( x )}{2 x} d x

Frequently Asked Questions (FAQ)

What is the solution for xy^'-2y=x^4-4 ?
The solution for xy^'-2y=x^4-4 is y=(x^4)/2+2+c_{1}x^2