What is the solution for y^{''}+0.2x^2-0.35x=0 ?

The solution for y^{''}+0.2x^2-0.35x=0 is y= 1/20 (-(x^4)/3+(7x^3)/6)+c_{1}x+c_{2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+0.2x^2-0.35x=0

y^{^{''}} + 0.2 x^{2} - 0.35 x = 0

y = \frac{1}{20} (- \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6}) + c_{1} x + c_{2}

Solution steps

y^{''} (x) + 0.2 x^{2} - 0.35 x = 0

Isolate

y^{''} (x) : y^{''} (x) = \frac{x ( - 4 x + 7 )}{20}

Solve

y^{''} (x) = \frac{x ( - 4 x + 7 )}{20} : y^{'} (x) = \frac{1}{20} (- \frac{4 x ^{3}}{3} + \frac{7 x ^{2}}{2}) + c_{1}

Solve

y^{'} (x) = \frac{1}{20} (- \frac{4 x ^{3}}{3} + \frac{7 x ^{2}}{2}) + c_{1} : y = \frac{1}{20} (- \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6}) + c_{1} x + c_{2}

y = \frac{1}{20} (- \frac{x ^{4}}{3} + \frac{7 x ^{3}}{6}) + c_{1} x + c_{2}

n = 1 \sum \infty \frac{1}{2 ^{3 n}}

f (x) = \frac{1}{2 x + 3}, at x = 2

\frac{d}{d x} (y cos (x))

cos (3 x^{2} + 4) \cdot 6 x

\int \frac{1}{3 x ^{2} + 6 x + 6} d x

What is the solution for y^{''}+0.2x^2-0.35x=0 ?
The solution for y^{''}+0.2x^2-0.35x=0 is y= 1/20 (-(x^4)/3+(7x^3)/6)+c_{1}x+c_{2}