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Solve the following equation with initial condition y(1) = 1 (4x + y)dx + (-x + y)dy = 0

User Astrit
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1 Answer

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Final Answer:

The solution to the given initial value problem is
\(y(x) = e^(-x)\).

Step-by-step explanation:

To solve the differential equation
\((4x + y)dx + (-x + y)dy = 0\) with the initial condition
\(y(1) = 1\), we can recognize it as a first-order linear differential equation. To solve it, we can use the method of separating variables.

The given equation can be rearranged as
\((4x + y)dx = (x - y)dy\). Now, separate variables by dividing both sides by
\((4x + y)\) and \((x - y)\):


\[(1)/(4x + y)dx = -(1)/(x - y)dy\]

Now, integrate both sides separately:


\[\int (1)/(4x + y)dx = -\int (1)/(x - y)dy\]

This leads to
\(\ln|4x + y| = -\ln|x - y| + C\), where
\(C\) is the constant of integration. Simplifying further, we get:


\[\ln|4x + y| + \ln|x - y| = C\]

Combine the logarithms using the properties of logarithms:


\[\ln|4x + y(x)||x - y(x)|| = C\]

Exponentiate both sides:


\[|4x + y(x)||x - y(x)| = e^C\]

Now, consider the initial condition
\(y(1) = 1\) to find the value of
\(C\). Substituting
\(x = 1\) and
\(y = 1\) into the equation, we get
\(5 = e^C\), so
\(C = \ln 5\).

Substitute
\(C\) back into the equation, simplify, and solve for
\(y(x)\) to obtain
\(y(x) = e^(-x)\), which is the solution to the initial value problem.

User XYShaoKang
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