How do you solve for xy'-y=3xy given y(1)=0?

I would write it as;xdydx=3xy+yxdydx=y(3x+1)dyy=3x+1xdxIntegrating;∫dyy=∫3x+1xdx∫dyy=∫(3+1x)dxln⁡(y)=3x+ln⁡(x)+cy=e3x+ln⁡(x)+cAnd finally;y=e3x×eln⁡(x)×ec=Axe3xwhere A=ecWe can now find the value of the constant A;0=Ae3A=0The particular solution to your equation is then:y=0xe3x=0

Step by step solution to "How do you solve for xy'-y=3xy given y(1)=0?"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Blaine Jimenez

Answered question

2022-04-11

How do you solve for xy'-y=3xy given y(1)=0?

Answer & Explanation

folytonr4qx

Beginner2022-04-12Added 14 answers

I would write it as;

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

x \frac{d y}{d x} = y (3 x + 1)

\frac{d y}{y} = \frac{3 x + 1}{x} d x

Integrating;

\int \frac{d y}{y} = \int \frac{3 x + 1}{x} d x

\int \frac{d y}{y} = \int (3 + \frac{1}{x}) d x

\ln (y) = 3 x + \ln (x) + c

y = e^{3 x + \ln (x) + c}

And finally;

y = e^{3 x} \times e^{\ln (x)} \times e^{c} = A x e^{3 x}

where

A = e^{c}

We can now find the value of the constant A;

0 = A e^{3}

A=0
The particular solution to your equation is then:

y = 0 x e^{3 x} = 0

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