{"id":5702,"date":"2021-09-29T18:37:05","date_gmt":"2021-09-29T13:07:05","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5702"},"modified":"2021-11-21T20:15:06","modified_gmt":"2021-11-21T14:45:06","slug":"solution-of-differential-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solution-of-differential-equation\/","title":{"rendered":"Solution of Differential Equation"},"content":{"rendered":"
Here you will learn how to find solution of differential equation i.e. general solution and particular solution with examples.<\/p>\n
Let’s begin –<\/p>\n
\nThe solution of differential equation is a relation between the variables involved which satisfies the differential equation.<\/p>\n
for example<\/strong>, y = \\(e^x\\) is a solution of the differential equations \\(dy\\over dx\\) = y.<\/p>\n<\/blockquote>\n
General Solution<\/strong><\/h4>\n
\nThe solution which contains as many as arbitrary constants as the order of the differential equations is called the general solution.<\/p>\n
for example<\/strong>, y = A cos x + B sin x is the general solution of the equation \\(d^2y\\over dx^2\\) + y = 0.<\/p>\n<\/blockquote>\n
Particular Solution<\/strong><\/h4>\n
\nThe solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equations is called a particular solution.<\/p>\n
for example<\/strong>, y = 3 cos x + 2 sin x is the particular solution of the equation \\(d^2y\\over dx^2\\) + y = 0.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Show that y = Ax\u00a0 + \\(B\\over x\\), x \\(\\ne\\) 0 is a solution of the differential equation \\(x^2\\)\\(d^2y\\over dx^2\\) + x\\(dy\\over dx\\) – y = 0<\/p>\n
Solution<\/span><\/strong> : We have,<\/p>\n
y = Ax\u00a0 + \\(B\\over x\\) = 0, x \\(\\ne\\) 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(i)<\/p>\n
Differentiating both sides with respect to x, we get<\/p>\n
\\(dy\\over dx\\) = A – \\(B\\over x^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(ii)<\/p>\n
Again differentiating with respect to x, we get<\/p>\n
\\(d^2y\\over dx^2\\) = \\(2B\\over x^3\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………..(iii)<\/p>\n
Substituting the values of y, \\(dy\\over dx\\) and \\(d^2y\\over dx^2\\) in \\(x^2\\)\\(d^2y\\over dx^2\\) + x\\(dy\\over dx\\) – y , we get<\/p>\n
\\(x^2\\)\\(d^2y\\over dx^2\\) + x\\(dy\\over dx\\) – y = \\(x^2\\)(\\(2B\\over x^3\\)) + x(A – \\(B\\over x^2\\)) – (Ax + \\(B\\over x\\))\u00a0<\/p>\n
= \\(2B\\over x\\) + Ax – \\(B\\over x\\) – Ax – \\(B\\over x\\) = 0<\/p>\n
Thus, the function y = Ax\u00a0 + \\(B\\over x\\) satisfies the given equation.<\/p>\n\n\n